Physics Problem: Work-Kinetic Frictional Force

[shawonna23]

A 280 N force is pulling an 80.0 kg refrigerator across a horizontal surface. The force acts at an angle of 18.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 6.00 m.
(a) Find the work done by the pulling force.
1.60*10^3 J
(b) Find the work done by the kinetic frictional force.
J

I got the answer for Part A right, but I can't seem to get the answer for part B. The answer i got for B is 940.8. I used the equation: W= -coefficient of kinetic friction*mg*distance but the answer is wrong. Am I using the wrong equation for Part B?

 

[Sirus]

Looks right. Remember friction does negative work here, so your final anwer needs to be negative (remember to use a positive value of g for this to work). Other than that, try different values of g, like 9.81, 9.8, 10, as these are all used by various texts.

 

[wais]

For part B, you are on the right track but you need to use the formula W = Ff * d, where Ff is the kinetic frictional force and d is the distance moved. The kinetic frictional force can be calculated using the formula Ff = μk * Fn, where μk is the coefficient of kinetic friction and Fn is the normal force, which is equal to the weight of the refrigerator in this case. So, your final equation should be W = μk * mg * d. Plugging in the values, we get W = 0.200 * 80.0 * 9.8 * 6.00 = 940.8 J. So, your answer is correct. Make sure to check your calculations again.