The drawing shows two boxes resting on frictionless ramps. One box is relatively

Question

The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a steep ramp. The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at A and allowed to slide down the ramps. The two boxes have masses of 12 and 40 kg. If A and B are 4.5 and 1.0 m, respectively, above the ground, determine the speed of (a) the lighter box and (b) the heavier box when each reaches B. (c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B?

(a) vB = Jm/smsNWm/s^2 (b) vB = m/s^2sWmm/sNJ (c) = The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a steep ramp. The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at A and allowed to slide down the ramps. The two boxes have masses of 12 and 40 kg. If A and B are 4.5 and 1.0 m, respectively, above the ground, determine the speed of (a) the lighter box and (b) the heavier box when each reaches B. (c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B? vB = Jm/smsNWm/s 2 vB = m/s 2sWmm/sNJ

Explanation / Answer

a)

As image not visible , consider steep makes angle theta with ground....

so,

acceleration = g*sin(theta)

so,

speed when it reach ground = Va = sqrt(2*a*s) = sqrt(2*9*81*sin(theta)*4.5) .....(jst put value of theta and calculate)

B)

consider steep makes angle alpha with ground....

so,

acceleration = g*sin(alpha)

so,

speed when it reach ground = Va = sqrt(2*a*s) = sqrt(2*9*81*sin(alpha)*1) .....(jst put value of alpha and calculate)

C)

ratio of Kinetic Enrgies = Mb*Vb^2/(Ma*Va^2) = 40*Vb^2/(12*Va^2) ...(put above value of Va and Vb)

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