Physics of work and pushing a refrigerator at an angle

A 2.40 x 102 N force is pulling an 85.0 kg refrigerator across a horizontal surface. The force acts at an angle of 20.0 above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 8.00 m. Determine the following:
a) The work done by the pulling force.
b) The normal force.
c) The force of friction.
d) The work done by the force of friction.

a) The work done by the pulling force is given by

W = Force * displacement
Here, the force is the component of the pulling force that is acting in the horizontal direction. So,

W = 2.4 * 10^2 * cos 20 * 8 = 1804.2 J

b) The normal force here has two parts,
i) the normal reaction of the refrigerator = mg = 85 * 9.81 = 833. 85 N
ii) the vertical component of the pulling force = 2.4 * 10^2 * sin20 = 82.10 N

Normal force = 833.85 + 82.1 = 915.95 N

c) The frictional force = μ * Normal force = 0.2 * 915.95 = 183.19 N

d) The frictional force acts in the opposite direction to the motion, and here, I am assigning the positive direction to the right, in the dirction of the horizontal component of the pullig force. So

Work done by frictional force = - (183.19 * 8) = - 1465.12 N

a) The work done by the pulling force = 240*(cos 20)*8.0 = 1804 J
b) The normal force = 85*9.8 - 240*sin 20 = 751N
c) The force of friction. = 751*0.2 = 150 N
d) The work done by the force of friction. = -150*8 = -1200 N