A 350 kg piano slides 3.6 m down a 34 incline and is kept from accelerating by a

Question

A 350 kg piano slides 3.6 m down a 34 incline and is kept from accelerating by a man who is pushing back on it parallel to the incline .The effective coefficient of kinetic friction is 0.38.

A. Calculate the force exerted by the man. Express your answer using two significant figures.

B. Calculate the work done by the man on the piano. Express your answer using two significant figures.

C. Calculate the work done by the friction force. Express your answer using two significant figures.

D. Calculate the net work done on the piano.

Explanation / Answer

here,

weight of the piano m = 350 kg ,

d = 3.6 m

= 34 degree

= 0.38

Normal = mg*cos

Grav acceleration = -m*g*sin

f rction = -*Normal = * mg*cos

a) F = 0 = F_man + F_grav + F_friction

F_man = -F_grav - F_friction

          = -mgsin +*mg*cos

F_man = 350*9.8 *sin(34) - 0.38*350*9.8*cos(34)

F_man = 837.43 N

b) F_man*d = 837.43*3.6m = 3014.748 J

c) the work done due to friction,

  wf = 0.38*350*9.8*cos(34)*3.6 m
      = 3890.043J

d)  the work done by the force of gravity,

Wg = -350*9.8*sin(34)*3.6 = -6904.91 J

netforce is 0, so work is zero = F_net*d,
        Wnet = 0 J
where F_net =0

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