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 8 and 32 kg. If A and B are 4.0 and 1.5 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 student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 63 kg, and the height of the water slide is 13.0 m. If the kinetic frictional force does -7.1 × 10³ J of work, how fast is the student going at the bottom of the slide?

Answer 1

1) Both the boxes will have same speed

from conservation of energy

mg(h_A - h_B) = 1/2 mv²

v_A = v_B = sqrt (2g(h_A - h_B)

v_A = v_B = sqrt (2*9.8*(4 - 1.5)

v_A = v_B = 7 m/s

ratio of the kinetic energy of the heavier box to that of the lighter box at B

1/2m_bv_b² / 1/2m_av_a² = 32/8 = 4

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P.E at top = mgh = 8026.2 J

Therefore,

K.E = 8026.2 - 7100

K.E = 926.2 J

tHEREFORE,

1/2mv² = 926.2

v = sqrt (2*926.2 / 63)

v = 5.422 m/s