How Is Work Calculated in the Presence of Kinetic Friction?

[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 have the answer for part a, but i keep getting the wrong answer for part b. Can someone please help me solve this problem?

 

[stunner5000pt]

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 have the answer for part a, but i keep getting the wrong answer for part b. Can someone please help me solve this problem?

for the work done by frictional force


first of all Ff = coeff(Mu) times Normal force (N)

so N + 280sin18 - 80g = 0

Ff = Mu N

Work done by friction is Ff times distance

 

[wais]

For part b, we can use the formula for work done by friction: W = Fdcosθ, where F is the force of friction, d is the distance, and θ is the angle between the force and the displacement.

First, we need to calculate the force of friction using the formula F = μN, where μ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the refrigerator, which is mg, where m is the mass and g is the acceleration due to gravity.

So, N = mg = (80.0 kg)(9.8 m/s^2) = 784 N

Then, F = μN = (0.200)(784 N) = 156.8 N

Now, we can plug in the values in the formula for work done by friction:

W = (156.8 N)(6.00 m)cos(18.0°) = (156.8 N)(6.00 m)(cos 18.0°) = 156.8 N(6.00 m)(0.951) = 893.376 J

Therefore, the work done by the kinetic frictional force is 893.376 J.