Question

A horizontal force of magnitude 32.1 N pushes a block of mass 3.91 kg across a floor where the coefficient of kinetic friction is 0.579. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 2.99 m across the floor? (b) During that displacement, the thermal energy of the block increases by 33.6 J. What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

Answer 1

Given Force ,F=32.1

Displacement ,d=2.99

(A) Finding work done

We know the formula for work done=F ×d

                                                        =32.1×2.99

Hence ,work done =95.979 N

(B) Calculating increase in thermal energy of floor

We know the total thermal energy formula is

    U=μ_km(g)(d)--Equation(1)

   Here μ_k=0.639

Mass , m=3.91 kg

g=9.81

d=2.99

Substituting all the values in Equation (1)

U=0.639×3.91×9.81×2.99

U=24.4852×2.99

U= 73.285 J

So, here increase in thermal energy =73.285 J -33.6 J                                                                    =39.685 J

(C) Finding increase in kinetic energy of block

So, hence increase in kinetic energy ∆k=w-u

                                                             =95.979 N-73.285 J

                                                                         

∆k=22.694 J