WORK

When a body has a displacement with magnitude s along a straight line while a constant force with magnitude F, direct along the same line. We define the work W done by the force on the body as


W=Fs................(1)
example 1

Steve is trying to impress Elaine with his new car, but the engine dies in the middle of an intersection. While Elaine steers, Styeve pushes the car 19 m to clear the intersection. If he pushes in the direction of motion with a constant force of 210 N (about 47 lb), how much work does he do on the car?

Solution
From W=Fs
=(210N)(19m)
=4.0 x 10^3 J
Answer.

Steve pushed the car in the direction he wanted it to go. What if he had pushed at an angle phi with the car displacement. Only the component in the direction of the car's motion, (210N)cos(phi), would be effective in moving the car. When the force F and the displacement s have different directions, we take the component of F in the dirction of the displacement s, and we define the work as the product of this component and the magnitude of the displacement. The component of F in the direction of s is Fcos(phi), so

W=(Fcos(phi))s.................(2)
or
W= F.s ........................(3)



brought to you by: University Physics ,Hugh D.Young Eighth edition


Example 2.

Consider a body with mass 0.2 kg move toward the right along smooth horizontal surface by a horizontal force F. The body has initial velocity of 2.8 m/s.
As a body move ,It slows down because of the horizontal friction force exerted on it by the horizontal surface.The displacement of a body is 1.0 m before coming to rest. What are the magnitude and direction of the friction force acting on it?

Solution
From the second law of motion with ay =0 we obtain

Fy=m.ay
or
N-W =0
when N is normal force, W is weight of a body

Hence N=W=mg=(0.2kg)(9.8 meters/sec^2)= 2 N


Suppose the body move toward in the +x direction, starting at the point x0=0 with the given initial velocity. We assume that the friction force f is constant. The acceleration is then constant also.Then we have

v^2=v0^2+2a(x-x0)
0=(2.8 m/s)^2+(2a)(1.0m),
a=-3.9 m/s^2)

The negative sign means that the acceleration is toward the left.
The friction force f then become

f=ma=(0.2kg)(-3.9 m/s^2)= -0.8 kg.m/s^2 =-0.8 N

Answer.


Example 3.
A block of mass 10.0 kg is to be raised from the bottom to the top of an incline 5.00 meters long and 3.00 meters off the ground at the top. Assuming friction less surface, how much work must be done by a force parallel to the incline pushing the block up at constant speed?


Solution

Define F = P= The magnitude of force pushing the block up the incline. Because the motion is not accelerated, the resultant force parallel to the plane must be zero. Thus

P-mgsin(theta) = 0,
or P=mgsin(theta)=(10.0 kg)(9.80 meters/sec^2)(3/5)= 58.8 N

When define s=d =displacement of a block.
Then the work done by P ,theta = 0 degree is

W=P.s=P.d =Pdcos(0)=Pd= (58.8 N)(5.00 meters)= 294 joules.






Answer.


Next: work done by variable force