A block of mass 3.3 kg is initially at rest on a wedge of mass 17 kg, height 0.3

Question

A block of mass 3.3 kg is initially at rest on a wedge of mass 17 kg, height  0.3m, and an incline angle = 35 degree as shown below.  There is no friction between the wedge and the floor,. Starting at the top of the incline, the block is released and slides toward the bottom of the wedge.  At the same time, the wedge " recoils" and slides some distance L to the right.  F ind L when the block has reached the bottom of the wedge.  Hint: The centre of masss of triangel of height h is a distance h/3 above the base on the triangle ( For simplicity, you may treat the block as a point mass)

A block of mass 3.3 kg is initially at rest on a wedge of mass 17 kg, height 0.30 m, and an incline angle theta = 35 degree as shown in the figure below. There is no friction between the wedge and the floor. Starting at the top of the incline, the block is released and slides toward the bottom of the wedge. At the same time, the wedge "recoils" and slides some distance L to the right. Find L when the block has reached the bottom of the wedge. Hint: The center of mass of a right triangle of height h is a distance h/3 above the base of the triangle. (For simplicity, you may treat the block as a point mass . ) L = cm

Explanation / Answer

conservation of energy

force along the inclined = mgsin35

horixontal component of force due to the normal reaftion of the small block = mgsin35cos35

mgsin35*h/sin35 = mgsin35*cos35*L

L = 3.3/sin35*cos35 = 3.3/(0.82) = 4.024m

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