7-0. An object of mass 10 kg is released at point A, slides down to the bottom o

Question

7-0. An object of mass 10 kg is released at point A, slides down to the bottom of a rough 30° incline, slides along a smooth horizontal surface and then collides with a spring compressing the spring by 75 cm. The spring constant is 500 N/m. 2.0 m 7. The speed at the bottom of the incline is The work of friction while the object was on the incline is '9·The spring recoils and sends the object back toward the incline. The speed of the object when it reaches the base of the incline is 10· The vertical distance it moves back up the incline is-

Explanation / Answer

7)

Using energy conservation principle at the bottom of the plank and when the block compresses the spring

mgh- W = 0.5*K*x^2

W = Work done by friction

W = mgh - 0.5Kx^2 = 10*9.8*2- 0.5*500*0.75*0.75 = 55.375 J (ans)

Kinetic energy of the ball when it reaches down = Spring potential energy

0.5mv^2= 0.5kx^2

5V^2 = 0.5*500*0.75*0.75

V = 5.30 m/sec (Ans)

9 ) When it reaches the base of the incline, speed will be same = 5.30 m/sec (Ans)

10) Using Energy conservation,

0.5*10*5.3*5.3 = 10*9.8*h

Vertical distance, h =1.43 m (ans)

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