A 39.0 kg box initially at rest is pushed 5.55 m along a rough, horizontal floor
ID: 2026022 • Letter: A
Question
A 39.0 kg box initially at rest is pushed 5.55 m along a rough, horizontal floor with a constant applied horizontal force of 125 N. If the coefficient of friction between box and floor is 0.300, find the following.(a) the work done by the applied force
____ J
(b) the increase in internal energy in the box-floor system due to friction
____ J
(c) the work done by the normal force
____ J
(d) the work done by the gravitational force
____ J
(e) the change in kinetic energy of the box
____ J
(f) the final speed of the box
____ m/s
Explanation / Answer
Given: Mass of the box , m = 39 kg moved distance , d = 5.5 m coefficient of friction between box and floor is ,= 0.300 applied horizontal force , F = 125 N initial velocity of the box , vi = 0 m/s _________________________________________________________ Solution: (a) work done by the applied force is , W = F.d = (125 N)(5.5 m ) = 687.5 Nm ___________________________________________________________ ___________________________________________________________ (b) increase in internal energy in the box-floor system due to frictionis nothing but work done by frictional force , Wfri = - fk d
= - (mg ) d Here , fk = mg is the frictional force on box - ve sign indicates frictional work is opposite to motion of the box Hence , Wfri = -(mg ) d = - (0.3)(39 kg)(9.8 m/s2)(5.5 m) = -630.63 Nm Hence , increase in internal energy in the box-floor system due to friction
is , 630.63 Nm (a) work done by the applied force is , W = F.d = (125 N)(5.5 m ) = 687.5 Nm ___________________________________________________________ ___________________________________________________________ (b) increase in internal energy in the box-floor system due to friction
is nothing but work done by frictional force , Wfri = - fk d
= - (mg ) d Here , fk = mg is the frictional force on box - ve sign indicates frictional work is opposite to motion of the box Hence , Wfri = -(mg ) d = - (0.3)(39 kg)(9.8 m/s2)(5.5 m) = -630.63 Nm Hence , increase in internal energy in the box-floor system due to friction
is , 630.63 Nm __________________________________________________________ __________________________________________________________ (c) when the box , there is no normal force acting on the box Hence , work done by the normal force is, WN= 0 Nm __________________________________________________________ __________________________________________________________ (d) when the box , there is no gravitational work is done on the box Hence , work done by the gravitational force is , Wg =0 Nm __________________________________________________________ ___________________________________________________________ (e) Acc. to work -energy theorem , change in kinetic energy of the box is equal to net work done on the box so , net work done on the box is , Wnet= W + (-Wfri ) + WN + Wg = 687.5 Nm - 630.63 Nm + 0 Nm + 0 Nm = 56.87 Nm Hence , change of kinetic enrgy of the box is , K.E = Wnet = 56.87 Nm or J _________________________________________________________
__________________________________________________________ (f)
but , change in kinetic energy , 1/2mvf2 - 1/2mvi2= Wnet so , 1/2mvf2 = Wnet = 56.87 J Hence , final speed of the box is , vf = [2(Wnet ) /m ] = [2(56.87 J) /(39 kg)] = 1.707 m/s