Forces and Work Done by Weight on a Ramp

[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.

 

[radou]

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

 

[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)

 

[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?

 

[radou]

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?

Yes, since cos(180) equals -1.

 

[arildno]

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?

Quite correct!
You could take the force component along the direction of motion; if that force component is parallell to the direction of motion, the angle between them is 0, whereas if it is anti-parallell, the angle is 180
The perpendicular part of the force would in any case vanish.