(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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Help with part c?

**Physics Forums | Science Articles, Homework Help, Discussion**