A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.(adsbygoogle = window.adsbygoogle || []).push({});

a) How much work is done by the worker on the crate?

b) How much work is fone by the floor on the crate?

c) What is the net work done on the crate?

--Okay, I understand that work equals force times the displacement times the cosin of theta. Howver, this "1.50 x 10^3 N crate" part is throwing me off! For part a, would I just use the formula and plug in 345 N or... see, I'm confused... someone care to explain? Thnks!

**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!

# Work and Force

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