An 66.0 N box of clothes is pulled 5.5 m up a 30.0° ramp by a force of 116 N that points along the ramp. If the coefficient of kinetic friction between the box and the ramp is 0.22, calculate the change in the box's kinetic energy.

1 answer

h = L*sin A = 5.5*sin30 = 2.75 m.

Fn = mg*Cos A = 66*Cos 30 = 57.16 N. =
Normal force = Force perpendicular to
the ramp.

Fk = u*Fn = 0.22 * 57.16 = 12.57 N. =
Force of kinetic friction.

KE + PE = mg*hmax-Fk*L
KE + PE = 66*2.75-12.57*5.5 = 112.4 J.
0 + PE = 112.4
PE = 112.4 J. at bottom of ramp.

KE + PE = 112.4 J. at top of ramp.
KE + 0 = 112.4
KE = 112.4 J.
KE(change) = 112.4 - 0 = 112.4 J.