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Solved Examples on Laws of Motion Question 1: A chain consisting of five links, each with mass 100 g, is lifted vertically with a constant acceleration of 2.50 m/s^{2}, as shown in below figure. Find, (a) The forces acting between adjacent links. (b) The force F exerted on the top link by the agent lifting the chain. (c) The net force on each link. Concept: Force acting on the object (F) is equal to the product of mas of the (m) and acceleration of the auto mobile (a). So, F_{net} = ma Weight of the automobile (W) is equal to the product of mass of the automobile (m) and acceleration due to gravity on the surface of the earth (g = 9.81 m/s^{2}). W = mg Solution: (a) The net force on each link is the same. To obtain the net force F_{net} on each link, substitute 0.100 kg for m and 2.50 m/s^{2} for a in the equation F_{net} = ma, F_{net} = ma =(0.100 kg)(2.50 m/s^{2}) =(0.250 kg.m/s^{2}) (1 N/1 kg.m/s^{2}) =0.250 N To obtain the weight W of the each link, substitute 0.100 kg for m and 9.81 m/s^{2} for g in the equation W = mg, W = mg =(0.100 kg)( 9.81 m/s^{2}) = (0.981 kg.m/s^{2}) (1 N/1 kg.m/s^{2}) =0.981 N On each link (except the top or bottom link) there is a weight W, an upward force U from the link above, and a downward force D from the link below. So, F_{net} = U-D-W Then, U= F_{net}+W+D =(0.250 N)+(0.981 N)+D =1.23 N+D For the bottom link, D=0 . So, U=1.23 N+D =1.23 N+0 =1.23 N For the link above, U = 1.23 N+1.23 N = 2.46 N For the next link above, U = 1.23 N+2.46 N = 3.69 N For the next link above, U = 1.23 N+3.69 N = 4.92 N From the above observation we conclude that, the force F exerted on the top link by the agent lifting the chain would be 4.92 N. (b) To obtain the upward force U for the top link, substitute 4.92 N for D in the equation U=1.23 N+D, U=1.23 N+D =1.23 N+4.92 N =6.15 N From the above observation we conclude that, the upward force U for the top link would be 6.15 N. (c) The net force on each link is the same. To obtain the net forces F_{net} acting between adjacent on each link, substitute 0.100 kg for m and 2.50 m/s^{2} for a in the equation F_{net} = ma, F_{net} = ma =(0.100 kg)(2.50 m/s^{2}) =(0.250 kg.m/s^{2}) (1 N/1 kg.m/s^{2}) =0.250 N From the above observation we conclude that, the forces acting between adjacent links would be 0.250 N. Question 2: (a) Two 10-lb weights are attached to a spring scale as shown in Fig. 1a. What is the reading of the scale? (b) A single 10-lb weight is attached to a spring scale which itself is attached to a wall, as shown in figure 1b. What is the reading of the scale? (Ignore the weight of the scale.) Solution: a) Let W_{1} is the weight of the left side block (W_{1} = 10-lb), a_{1} is the upward acceleration of the block and m_{1} is the mass of the block of the left hand side. The upward force due to the upward acceleration a_{1} will be, F_{1}= m_{1}a_{1} …… (1) If T is the tension in the upward direction on the left side block (shown in the free body diagram), then the net force will be, m_{1}a_{1}= W_{1}-T …… (2) Let W_{2} is the weight of the right side block (W_{2} = 10-lb), a_{2} is the downward acceleration of the block and m_{2} is the mass of the block of the right hand side. The downward force due to the downward acceleration a_{2} on the right side block will be, F_{2}= m_{2}a_{2} …… (3) If T is the tension in the upward direction on the left side block (shown in the free body diagram), then the net force will be, m_{2}a_{2}= W_{2}-T …… (4) FBD:- Subtracting equation (4) from equation (2), m_{1}a_{1}- m_{2}a_{2} = W_{1}-T –T+ W_{2} …… (5) Since, W_{1}= W_{2}=10-lb and there is no relative motion, the left side of the equation (5) will be zero. So, W_{1}-T –T+ W_{2}= 0 2T = (W_{1} + W_{2}) = (10-lb+10-lb) So, T = 20-lb/2 =10-lb …… (6) From equation (6) we observed that, the reading of the scale will be 10-lb. b) When a single 10-lb weight (W=10-lb) is attached to a spring scale which itself is attached to a wall, the net exerted force (F_{net}) on the block will be equal to weight of the block (W) minus tension exerted on the block (T). So, F_{net} = W-T …… (7) Since the weight is attached to a spring scale which itself is attached to a wall, therefore the net force will be zero (F_{net}=0). Putting the value of F_{net} in equation (7), we get, W-T=0 T = W = 10-lb …… (8) From equation (8) we observed that, the reading of the scale will be 10-lb. Question 3:- A 1200-kg car is being towed up an 18^{°} incline by means of a rope attached to the rear of a truck. The rope makes an angle of 27^{°} with the incline. What is the greatest distance that the car can be towed in the first 7.5 s starting from rest if the rope has a breaking strength of 4.6 kN? Ignore all the resistive forces on the car. See the low figure. Concept: The figure below shows the car of mass M towed by another car through a rope of breaking strength T. Assume that the x axis is along the length of the ramp and y axis is perpendicular to it such that the tension in the rope along the x axis is whereas the tension in the rope along the y axis is . Apply Newton’s second law of motion along the x axis, T cos (27^{°}) – Mg sin (18°) = Ma_{x} Here a_{x} is the acceleration of the car of mass M along the x axis and g is the acceleration due to gravity. a_{x} = T cos (27^{°}) – Mg sin (18°)/M From this acceleration, one can calculate the distance travelled by the car in time t using relation, x = ½ a_{x}t^{2} Solution: Substitute 1200 kg for M, 9.8 m/s^{2} for g and 4.6 kN for T in equation, a_{x} = [T cos (27^{°}) – Mg sin (18^{°})]/M a_{x} = [(4.6 kN) (10^{3} N/1 kN) cos (27^{°}) – (1200 kg) (9.81 m/s^{2})sin (18^{°})]/[1200 kg] = {(4098.6 N) [(1 kg.m/s^{2})/1 N]/[1200 kg]}-3.02 m/s^{2} = 3.41 m/s^{2} - 3.02 m/s^{2} = 0.38 m/s^{2} Therefore, the acceleration of the car towed up by another car with a rope of breaking strength 4.6 kN is 0.38 m/s^{2}. Substitute 0.38 m/s^{2} for a_{x} and 7.5 s for t in equation x = ½ a_{x}t^{2}, x = ½ (0.38 m/s^{2}) (7.5 s)^{2} = 10.8 m Therefore, the distance by which the car can move in 7.5 s is 10.8 m. Question 4:- Someone exerts a force F directly up on the axle of the pulley shown in below figure. Consider the pulley and string to be mass less and the bearing frictionless. Two objects, m_{1} of mass 1.2 kg and m_{2} of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The object m_{2} is in contact with the floor. (a) What is the largest value of force may have so that m_{2} will remain at rest on the floor? (b) What is the tension in the string if the upward force F is 110 N? (c) With the tension determined in part (b), what is the acceleration of m_{1}? Concept: The figure below shows the forces acting in the block-pulley system: It is important to note that the tension pulling the pulley down is 2T whereas the force pushing it upwards is F. Therefore, the block with mass m_{2} will stay on floor when F ≤ 2T, and the tension in the string T is equal to the weight of the block on floor. The net force on block with mass m_{1} is given as: F_{net} = F-m_{1}g Thus, the acceleration of block with mass m_{1} is: a = (T-m_{1}g)/m_{1} Solution: (a) Substitute 1.9 kg for m_{2}, and 9.8 m/s^{2} for g in equation T = m_{2}g, T = (1.9 kg) (9.8 m/s^{2}) = (18.62 kg. m/s^{2}) [1 N/(1kg. m/s^{2})] = 18.62 N The maximum force F for which the block stays on the floor is F= 2T. Substitute 18.62 N for T in equation F = 2T, F = 2 (18.62 N) = 37.24 N Round off to two significant figures F = 37 N Therefore, the maximum force F for which block with mass m_{2} stays on floor is 37 N. (b) The tension in the string is related to upward force as: T = F/2 Substitute 110 N for F, T = F/2 = 110 N/2 = 55 N Therefore, the tension in the string corresponding to upward force of 110 N is 55 N. (c) Substitute 55 N for T, 1.2 kg for m_{1} and 9.8 m/s^{2} for g in equation a = T-m_{1}g/m_{1} a = [55 N- (1.2 kg) (9.8 m/s^{2})]/1.2 kg = {(55 N) [(1 kg.m/s^{2})/1 N]- (11.76 kg.m/s^{2})}/1.2 kg = 36 m/s^{2} Therefore, the acceleration of the block with mass m_{1} is 36 m/s^{2}. Question 5:- The coefficient of static friction between the tires of a car and a dry road is 0.62. The mass of the car is 1500 kg. What maximum braking force is obtainable (a) on a level road and (b) on an 8.6°downgrade? Concept: Assume that the mass of the car is m and the coefficient of friction between the tires of the car and the road is µ_{s}. The magnitude of maximum braking force obtainable on the level road is equal to the magnitude of the static frictional force between the tires and the road. The magnitude of the frictional force is, f = µ_{s} N Here N is the magnitude of normal force on the car. (a) The car is in vertical equilibrium on the level road therefore the normal force on the car should be equal to its weight (mg) in magnitude, and must act in opposite direction, that is N = mg Substitute N = mg in equation f = µ_{s} N , f = µ_{s} N = µ_{s}mg Therefore, the magnitude of maximum obtainable braking force is . (b) When the car is on the inclined road, the normal force is not equal to the weight of the car but to its component mg cosθ, that is N = mg cosθ Here θ is the angle of inclination of the road. Substitute N = mg cosθ in equation f = µ_{s} N, f = µ_{s} N = µ_{s} mg cosθ Therefore, the magnitude of maximum obtainable braking force on the inclined road is µ_{s} mg cosθ. Solution: (a) Substitute 0.62 for µ_{s}, 1500 kg for m and 9.81 m/s^{2} for g in equation F = µ_{s} mg, F = µ_{s} mg = (0.62) (1500 kg) (9.81 m/s^{2}) = (9123.3 kg.m/s^{2}) [1 N/(1 kg.m/s^{2})] = 9123. 3 N Round off to four significant figures, F = 9123 N Therefore, the maximum obtainable braking force on the level road is 9123 N. (b) Substitute 0.62 for µ_{s}, 1500 kg for m, 8.6^{°} for θ and 9.81 m/s^{2} for g in equation F = µ_{s} mg cosθ, F = µ_{s} mg cosθ = (0.62) (1500 kg) (9.81 m/s^{2}) cos (8.6^{°}) = (9020 kg. m/s^{2}) [1 N/(1 kg.m/s^{2})] = 9020 N Therefore, the maximum obtainable braking force on the inclined road is 9020 N. Question 6:- A railroad flat car of weight W can roll without friction along a straight line track. Initially, a man of weight w is standing on the car, which is moving to the right with speed v_{0}. What is the change in velocity of the car if the man runs to the left (below figure) so that his speed relative to the car is v_{rel} just before he jumps off at the left end? Concept: Momentum of the body p is equal to the mass of the body m times velocity of the body v. So, p = mv In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system. Consider the initial momentum of the man is p_{i,m}, initial momentum of the car is p_{i,c}, final momentum of the man is p_{f,m} and final momentum of the car is p_{f,c}. Initial momentum of the man is equal to the initial momentum of the car. So, initial momentum of the man will be, p_{i,m} = m_{m}v_{i,c} and final momentum of the man p_{f,m} will be, p_{f,m} = m_{m}(v_{f,c} – v_{rel}) Here, speed of the man relative to the car is v_{rel} . Again, p_{f,c} = m_{c}v_{f,c} p_{i,c} =...
Question 1: A chain consisting of five links, each with mass 100 g, is lifted vertically with a constant acceleration of 2.50 m/s^{2}, as shown in below figure. Find,
(a) The forces acting between adjacent links.
(b) The force F exerted on the top link by the agent lifting the chain.
(c) The net force on each link.
Concept:
Force acting on the object (F) is equal to the product of mas of the (m) and acceleration of the auto mobile (a).
So, F_{net} = ma
Weight of the automobile (W) is equal to the product of mass of the automobile (m) and acceleration due to gravity on the surface of the earth (g = 9.81 m/s^{2}).
W = mg
Solution:
(a) The net force on each link is the same. To obtain the net force F_{net} on each link, substitute 0.100 kg for m and 2.50 m/s^{2} for a in the equation F_{net} = ma,
F_{net} = ma
=(0.100 kg)(2.50 m/s^{2})
=(0.250 kg.m/s^{2}) (1 N/1 kg.m/s^{2})
=0.250 N
To obtain the weight W of the each link, substitute 0.100 kg for m and 9.81 m/s^{2} for g in the equation W = mg,
=(0.100 kg)( 9.81 m/s^{2})
= (0.981 kg.m/s^{2}) (1 N/1 kg.m/s^{2})
=0.981 N
On each link (except the top or bottom link) there is a weight W, an upward force U from the link above, and a downward force D from the link below.
So, F_{net} = U-D-W
Then, U= F_{net}+W+D
=(0.250 N)+(0.981 N)+D
=1.23 N+D
For the bottom link, D=0 .
So, U=1.23 N+D
=1.23 N+0
=1.23 N
For the link above,
U = 1.23 N+1.23 N
= 2.46 N
For the next link above,
U = 1.23 N+2.46 N
= 3.69 N
U = 1.23 N+3.69 N
= 4.92 N
From the above observation we conclude that, the force F exerted on the top link by the agent lifting the chain would be 4.92 N.
(b) To obtain the upward force U for the top link, substitute 4.92 N for D in the equation U=1.23 N+D,
U=1.23 N+D
=1.23 N+4.92 N
=6.15 N
From the above observation we conclude that, the upward force U for the top link would be 6.15 N.
(c) The net force on each link is the same. To obtain the net forces F_{net} acting between adjacent on each link, substitute 0.100 kg for m and 2.50 m/s^{2} for a in the equation F_{net} = ma,
From the above observation we conclude that, the forces acting between adjacent links would be 0.250 N.
Question 2: (a) Two 10-lb weights are attached to a spring scale as shown in Fig. 1a. What is the reading of the scale?
(b) A single 10-lb weight is attached to a spring scale which itself is attached to a wall, as shown in figure 1b. What is the reading of the scale? (Ignore the weight of the scale.)
a) Let W_{1} is the weight of the left side block (W_{1} = 10-lb), a_{1} is the upward acceleration of the block and m_{1} is the mass of the block of the left hand side.
The upward force due to the upward acceleration a_{1} will be,
F_{1}= m_{1}a_{1} …… (1)
If T is the tension in the upward direction on the left side block (shown in the free body diagram), then the net force will be,
m_{1}a_{1}= W_{1}-T …… (2)
Let W_{2} is the weight of the right side block (W_{2} = 10-lb), a_{2} is the downward acceleration of the block and m_{2} is the mass of the block of the right hand side.
The downward force due to the downward acceleration a_{2} on the right side block will be,
F_{2}= m_{2}a_{2} …… (3)
m_{2}a_{2}= W_{2}-T …… (4)
FBD:-
Subtracting equation (4) from equation (2),
m_{1}a_{1}- m_{2}a_{2} = W_{1}-T –T+ W_{2} …… (5)
Since, W_{1}= W_{2}=10-lb and there is no relative motion, the left side of the equation (5) will be zero.
So, W_{1}-T –T+ W_{2}= 0
2T = (W_{1} + W_{2})
= (10-lb+10-lb)
So, T = 20-lb/2
=10-lb …… (6)
From equation (6) we observed that, the reading of the scale will be 10-lb.
b) When a single 10-lb weight (W=10-lb) is attached to a spring scale which itself is attached to a wall, the net exerted force (F_{net}) on the block will be equal to weight of the block (W) minus tension exerted on the block (T).
So, F_{net} = W-T …… (7)
Since the weight is attached to a spring scale which itself is attached to a wall, therefore the net force will be zero (F_{net}=0).
Putting the value of F_{net} in equation (7), we get,
W-T=0
T = W
= 10-lb …… (8)
From equation (8) we observed that, the reading of the scale will be 10-lb.
Question 3:- A 1200-kg car is being towed up an 18^{°} incline by means of a rope attached to the rear of a truck. The rope makes an angle of 27^{°} with the incline. What is the greatest distance that the car can be towed in the first 7.5 s starting from rest if the rope has a breaking strength of 4.6 kN? Ignore all the resistive forces on the car. See the low figure.
The figure below shows the car of mass M towed by another car through a rope of breaking strength T.
Assume that the x axis is along the length of the ramp and y axis is perpendicular to it such that the tension in the rope along the x axis is whereas the tension in the rope along the y axis is .
Apply Newton’s second law of motion along the x axis,
T cos (27^{°}) – Mg sin (18°) = Ma_{x}
Here a_{x} is the acceleration of the car of mass M along the x axis and g is the acceleration due to gravity.
a_{x} = T cos (27^{°}) – Mg sin (18°)/M
From this acceleration, one can calculate the distance travelled by the car in time t using relation,
x = ½ a_{x}t^{2}
Substitute 1200 kg for M, 9.8 m/s^{2} for g and 4.6 kN for T in equation,
a_{x} = [T cos (27^{°}) – Mg sin (18^{°})]/M
a_{x} = [(4.6 kN) (10^{3} N/1 kN) cos (27^{°}) – (1200 kg) (9.81 m/s^{2})sin (18^{°})]/[1200 kg]
= {(4098.6 N) [(1 kg.m/s^{2})/1 N]/[1200 kg]}-3.02 m/s^{2}
= 3.41 m/s^{2} - 3.02 m/s^{2}
= 0.38 m/s^{2}
Therefore, the acceleration of the car towed up by another car with a rope of breaking strength 4.6 kN is 0.38 m/s^{2}.
Substitute 0.38 m/s^{2} for a_{x} and 7.5 s for t in equation x = ½ a_{x}t^{2},
x = ½ (0.38 m/s^{2}) (7.5 s)^{2}
= 10.8 m
Therefore, the distance by which the car can move in 7.5 s is 10.8 m.
Question 4:- Someone exerts a force F directly up on the axle of the pulley shown in below figure. Consider the pulley and string to be mass less and the bearing frictionless. Two objects, m_{1} of mass 1.2 kg and m_{2} of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The object m_{2} is in contact with the floor. (a) What is the largest value of force may have so that m_{2} will remain at rest on the floor? (b) What is the tension in the string if the upward force F is 110 N? (c) With the tension determined in part (b), what is the acceleration of m_{1}?
The figure below shows the forces acting in the block-pulley system:
It is important to note that the tension pulling the pulley down is 2T whereas the force pushing it upwards is F. Therefore, the block with mass m_{2} will stay on floor when
F ≤ 2T, and the tension in the string T is equal to the weight of the block on floor.
The net force on block with mass m_{1} is given as:
F_{net} = F-m_{1}g
Thus, the acceleration of block with mass m_{1} is:
a = (T-m_{1}g)/m_{1}
(a) Substitute 1.9 kg for m_{2}, and 9.8 m/s^{2} for g in equation T = m_{2}g,
T = (1.9 kg) (9.8 m/s^{2})
= (18.62 kg. m/s^{2}) [1 N/(1kg. m/s^{2})]
= 18.62 N
The maximum force F for which the block stays on the floor is F= 2T.
Substitute 18.62 N for T in equation F = 2T,
F = 2 (18.62 N)
= 37.24 N
Round off to two significant figures
F = 37 N
Therefore, the maximum force F for which block with mass m_{2} stays on floor is 37 N.
(b) The tension in the string is related to upward force as:
T = F/2
Substitute 110 N for F,
= 110 N/2
= 55 N
Therefore, the tension in the string corresponding to upward force of 110 N is 55 N.
(c) Substitute 55 N for T, 1.2 kg for m_{1} and 9.8 m/s^{2} for g in equation a = T-m_{1}g/m_{1}
a = [55 N- (1.2 kg) (9.8 m/s^{2})]/1.2 kg
= {(55 N) [(1 kg.m/s^{2})/1 N]- (11.76 kg.m/s^{2})}/1.2 kg
= 36 m/s^{2}
Therefore, the acceleration of the block with mass m_{1} is 36 m/s^{2}.
Question 5:- The coefficient of static friction between the tires of a car and a dry road is 0.62. The mass of the car is 1500 kg. What maximum braking force is obtainable
(a) on a level road and
(b) on an 8.6°downgrade?
Assume that the mass of the car is m and the coefficient of friction between the tires of the car and the road is µ_{s}.
The magnitude of maximum braking force obtainable on the level road is equal to the magnitude of the static frictional force between the tires and the road. The magnitude of the frictional force is,
f = µ_{s} N
Here N is the magnitude of normal force on the car.
(a) The car is in vertical equilibrium on the level road therefore the normal force on the car should be equal to its weight (mg) in magnitude, and must act in opposite direction, that is
N = mg
Substitute N = mg in equation f = µ_{s} N ,
= µ_{s}mg
Therefore, the magnitude of maximum obtainable braking force is .
(b) When the car is on the inclined road, the normal force is not equal to the weight of the car but to its component mg cosθ, that is
N = mg cosθ
Here θ is the angle of inclination of the road.
Substitute N = mg cosθ in equation f = µ_{s} N,
= µ_{s} mg cosθ
Therefore, the magnitude of maximum obtainable braking force on the inclined road is µ_{s} mg cosθ.
(a) Substitute 0.62 for µ_{s}, 1500 kg for m and 9.81 m/s^{2} for g in equation F = µ_{s} mg,
F = µ_{s} mg
= (0.62) (1500 kg) (9.81 m/s^{2})
= (9123.3 kg.m/s^{2}) [1 N/(1 kg.m/s^{2})]
= 9123. 3 N
Round off to four significant figures,
F = 9123 N
Therefore, the maximum obtainable braking force on the level road is 9123 N.
(b) Substitute 0.62 for µ_{s}, 1500 kg for m, 8.6^{°} for θ and 9.81 m/s^{2} for g in equation
F = µ_{s} mg cosθ,
F = µ_{s} mg cosθ
= (0.62) (1500 kg) (9.81 m/s^{2}) cos (8.6^{°})
= (9020 kg. m/s^{2}) [1 N/(1 kg.m/s^{2})]
= 9020 N
Therefore, the maximum obtainable braking force on the inclined road is 9020 N.
Question 6:- A railroad flat car of weight W can roll without friction along a straight line track. Initially, a man of weight w is standing on the car, which is moving to the right with speed v_{0}. What is the change in velocity of the car if the man runs to the left (below figure) so that his speed relative to the car is v_{rel} just before he jumps off at the left end?
Momentum of the body p is equal to the mass of the body m times velocity of the body v.
So, p = mv
In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system.
Consider the initial momentum of the man is p_{i,m}, initial momentum of the car is p_{i,c}, final momentum of the man is p_{f,m} and final momentum of the car is p_{f,c}.
Initial momentum of the man is equal to the initial momentum of the car.
So, initial momentum of the man will be,
p_{i,m} = m_{m}v_{i,c}
and final momentum of the man p_{f,m} will be,
p_{f,m} = m_{m}(v_{f,c} – v_{rel})
Here, speed of the man relative to the car is v_{rel} .
Again, p_{f,c} = m_{c}v_{f,c}
p_{i,c} =...
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