Question

A 10 kg box is hit by a truck and slides across concrete. After being hit by the truck the box has an initial velocity of 6 m/s and takes 5 m to come to a stop. If the box keeps up by 10° what is the specific heat of the box and the coefficient of friction between the box and the concrete? (If possible please include a free body diagram!)

Answer 1

W = work done by frictional force in bringing the box to a stop = Q = Heat gained by the box

m = mass of the box = 10 kg

c = specific heat of box

T = Change in temperature = 10

v = initial speed of the box = 6 m/s

Heat gained by the box is same as the kinetic energy lost , hence

W = (0.5) m v²

Q = (0.5) m v²

m c T = (0.5) m v²

c T = (0.5) v²

c (10) = (0.5) 6²

c = 1.8 J/(Kg-C)

N = normal force on the box

Normal force on the box is given as

N = mg

N = (10)(9.8)

N = 98 N

v_o = initial velocity of the box after being hit by truck = 6 m/s

v_f = final velocity of the box = 0 m/s

a = acceleration of the box

d = stopping distance = 5

using the kinematics equation

v_f² = v_o² + 2 a d

0² = 6² + 2 a (5)

a = - 3.6 m/s²

magnitude of frictional force is given as

f = ma

f = (10) (3.6)

f = 36 N

Coefficient of kinetic friction is given as

= f/N

= 36/98

= 0.37