# Forces and Work Done by Weight on a Ramp

1. Oct 7, 2006

### Soaring Crane

1) A constant external force P=170 N at 30 degrees above the horizontal is applied to a 20 kg box, which is on a rough horizontal surface. The force pushes the box a distance of 8.0 m, in a time interval 0f 6.0 s, and the speed changes from v1 = 0.3 m/s to v2 = 2.5 m/s. The work done by external force P is closest to:

a.680 J--------b. 1060 J----------------c. 940 J----------d. 810 J-----------e. 1180 J

Well F_P = 170*cos (30), then W = 170*cos (30)*8.0 m = 1178 J??

2) A 300 kg crate is on a rough surface inclined at 30 degrees. A constant external force P = 2400 N is applied horizontally to the crate. The force pushes the crate a distance of 3.0 m up the incline, in a time interval of 9.0 s, and the velocity changes from v1 = 0.7 m/s to v2 = 2.9 m/s. The work done by the weight is closest to:

a.0--------b.-4400 J----------c. 1400 J-------------d.4400 J-----e. –1400 J

The weight has a component in the direction of displacement, and it’s mg*sin(30).
W = mg*sin(30)*d = (300 kg)*(9.80 m/s^2)*sin(30)*3.0 m = 4410 J

But is the work done positive? Is this even the correct setup that I have?

Thanks.

2. Oct 7, 2006

The work done is positive if its 'direction' is the same as the direction of motion.

3. Oct 7, 2006

### arildno

In b) you work AGAINST gravity, i.e its work should be negative.
The reason for you getting it wrong, is that mgsin(30) points DOWNWARDS, whereas the directed the distance the crate moves is UPWARDS.

In order to get this right, set your directed distance as a VECTOR:
$$\vec{D}=d(\cos(30)\vec{i}+\sin(30)\vec{j}), \vec{W}=-mg\vec{j}, d=3.0 m$$
Thus the scalar W, work, becomes:
$$W=\vec{D}\cdot\vec{W}=-dmg\sin(30)$$

Last edited: Oct 7, 2006
4. Oct 7, 2006

### Soaring Crane

For #2, would it also be correct to multiply by cos(180) for work since this is the angle of the weight's component from the direction of displacement?

5. Oct 7, 2006