. A 70.0-kg (including the passenger) sled is subject to a net force of 30.0 N p

Question

. A 70.0-kg (including the passenger) sled is subject to a net force of 30.0 N pushing in the direction of the sled’s motion as it is moving over a horizontal surface for a distance of 12.0 m after having started from rest. At this point the sled is released as it starts down a 10.0° incline. However, the snow is not very deep, and the sled stops after having moved an additional 45.0 m. What is the work done by friction while the sled is on the incline?

A. –31381.44 J

B. –30661.44 J

C. –5829.92 J

D. –5469.92 J

E. –5109.92 J

Explanation / Answer

here,

mass , m = 70 kg

force , F = 30 N

the kinetic energy gained by the sled , KE = work done by force

KE = F * s = 360 J

the work done by friction , ff = - ( initial kinetic energy + potential energy lost)

ff = - ( 360 + m * g * ( l * sin(theta)))

ff = - ( 360 + 70 * 9.81 * ( 45 * sin(10)))

ff = - 5829.92 J

the work done by frictional force is C) - 5829.92 J

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