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 13 and 34 kg. If A and B are h_A = 4.3 and h_B = 1.6 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?

Answer 1

Answer 2

Your link is secured behind a login page so worthless for most of us.
I think I have the picture anyway.

At B both crates will have changed vertical position by a distance h of 1.5 m
The change in kinetic energy will equal the change in potential energy
PE = KE
mgh =

Answer 3

At B both crates will have changed vertical position by a distance h of 1.6 m
The change in kinetic energy will equal the change in potential energy
PE = KE
mgh =

Answer 4

(a) by conservation of energy,

M*g*4.3=.5*M*v^2

v=9.185m/s

(b) Mb*g*1.6=.5*Mb*v^2

v=5.602m/s

(c) ratio=K.E.b/K.E.b

   = 34*(5.602)^2/(13*(9.1850^2))

=.9728

Answer 5

Because we are neglecting friction heavy and light objects will fall at the same rate.

To solve this problem we can equate potential energy to kinetic energy and solve for speed:

so mgh = .5mv^2

The masses kill each other (see the first sentence) and solving for v we get: v = sqrt[2gh]. In this case the hight is 4.3 - 1.6 = 2.7 so v = sqrt[2(9.8)(2.7)] = 7.27 m/s. This is the answer to 'a' and 'b'.

For question 'c' the kinetic energy is .5mv^2 so the ratio of the kinetic energies is just the ratio of the masses: 34/13 = 2.61

Answer 6

Take help

Answer 7

Since the ramps are frictionless, entire P.E shall be turning into K.E

So,

1/2 m_av_a² = m_agh_a

Thus, v_a = sqrt (2gh_a) = 9.18 m/s

similarly,The heavier box at B shall have K.E developed from loss of P. E

So, m_b g (4.3-1.6) = 0.5 m_b v_b²

Thus, v_b = 7.27 m/s

Ratio of K.E at B shall be the just the ratio of their weights because, irrespective of their weights, the speed of both the boxes at B when released at rest from A shall be the same.

Thus, ratio of KE = 34 / 13 = 2.61

Hope this helps.