(adsbygoogle = window.adsbygoogle || []).push({}); help with part c??

A worker of a moving company places a 252 kg trunk on a piece of carpeting and slides it across the floor at constant velocity by exerting a horizontal force of 425 N on the trunk.

a) what is the coeffient of kinetic friction

Fn = mg

= 252 kg x 9.8 N/kg

= 2469.6 N

kinetic coefficient equals 425 N/2469.6 N = 0.17

b) what happens to the coefficient of kinetic friction if another 56 kg trunk is splaced on top of the 252 kg trunk?

Fn = mg

=(308 kg) (9.8 N/kg)

=3018.4

kinetic coefficient equals 425 N/3018.4 N = 0.14

c) what horizontal force must the mover apply to move the combination of the two trunks at constant velocity

the answer is 5.2 x 10^2 N any help would be appreciated as to how to solve c

thank-you

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# Homework Help: Help with part c?

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