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Grade: 12
`        consider the inclined plane with an angle of 67 degree. The spring constant has a value of 1850 N/m. A 10kg block is pushed against the spring compressing it to 40 cm. The block is then released and the spring expands pushing the block up the plane. the coefficient of kinetic friction between the block and the plane is 0.5. what is the normal force and frictional force acting on the system? how far up the plane will the block slide before coming to stop? what is the change in KE?`
2 years ago

Answers : (1)

Sanju
106 Points
```								Normal force = mg cos 67 = 10*9.81*cos67 = 38.33N	Frictional force = coeff. of kinetic friction * normal force = 0.5*38.33 = 19.165N	initial (KE+Gravitational PE+ Spring KE)-W = final (KE+Gravitational PE+ Spring KE)......this eqn gets deduced to this form: [spring PE – W = Grav. PE]......i.e, [(0.5 k(x)^2) – (Fd) = mgh]....Therefore d = [(0.5 k(x)^2) / (F+mg sin 67)]...Plugging the values give d = 1.35 m...So it will move 1.35 m up the plane...	Change in KE is zero as initial and final velocities are zero.
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2 years ago
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