Work done by gravitational force (new problem)

[momoneedsphysicshelp]

Homework Statement

A block M, with mass of 4 kg, is released from rest and slides down a ramp that is angled at 28° above the horizontal (from the bottom of the ramp). The length of the ramp, d, is 3 meters. How much work is done by the gravitational force as the block slides down the ramp?

Relevant Equations

F=ma, W= F*d*cos(theta)

first to find the force by gravity, it would be
4 kg * 9.8 m/s = 39.2 N
then solve for work using W= F*d*cos(theta)
W = 39.2 N * 3 m * cos 28
= 103.83 J

My confusion is do I use sine or cosine and what angle do I use, the actual angle of the incline or the angle between the mass and the direction of gravity?

Please help if you know, Thanks.

 

[PeroK]

Homework Statement:: A block M, with mass of 4 kg, is released from rest and slides down a ramp that is angled at 28° above the horizontal (from the bottom of the ramp). The length of the ramp, d, is 3 meters. How much work is done by the gravitational force as the block slides down the ramp?
Relevant Equations:: F=ma, W= F*d*cos(theta)

first to find the force by gravity, it would be
4 kg * 9.8 m/s = 39.2 N
then solve for work using W= F*d*cos(theta)
W = 39.2 N * 3 m * cos 28
= 103.83 J

My confusion is do I use sine or cosine and what angle do I use, the actual angle of the incline or the angle between the mass and the direction of gravity?

Please help if you know, Thanks.

What if $\theta = 0$?

 

[PeroK]

Relevant Equations:: F=ma, W= F*d*cos(theta)

This formula is practically useless unless you know what $\theta$ means.

 

[momoneedsphysicshelp]

What if $\theta = 0$?

theta would be 28?

 

[PeroK]

theta would be 28?

I meant, what if you changed the problem so that $\theta = 0$. Instead of $\theta = 28°$. Would that help you work out whether you need $\sin \theta$ or $\cos \theta$?

 

[haruspex]

My confusion is do I use sine or cosine and what angle do I use, the actual angle of the incline or the angle between the mass and the direction of gravity?

I think you mean the angle between the displacement and the direction of gravity.
On your previous thread, you declined to answer my question. Please answer this time:
What is that angle, numerically?

 

[momoneedsphysicshelp]

I think you mean the angle between the displacement and the direction of gravity.
On your previous thread, you declined to answer my question. Please answer this time:
What is that angle, numerically?

It would be the complementary angle to 28 degrees, so 62 degrees

 

[momoneedsphysicshelp]

I tried the problem again using sin instead of cosine.
work done= force x distance moved in the direction of the force. The vertical height fallen is 3sin28
so work = change in gravitational potential energy = mgh = 4 x 9.81 x 3sin28 = 55J
does this seem more right than my previous answer?

 

[jbriggs444]

work done= force x distance moved in the direction of the force. The vertical height fallen is 3sin28
so work = change in gravitational potential energy = mgh = 4 x 9.81 x 3sin28 = 55J
does this seem more right than my previous answer?

Yes, this would be a correct approach and a good insight: work = force times distance moved in the direction of the force.

Equivalently, work is distance moved times the component of force in the direction of the distance moved.

There is also the formula: The dot product is given by the product of the magnitudes and the [as it turns out] cosine of the angle between the two.

I've never bothered memorizing whether it's sine or cosine for the dot product. Instead, I use the approach that @PeroK was alluding to. Consider what happens for small angles. And for near right angles.

Or, take the easy way out: Google "dot product sine or cosine".

 

[haruspex]

It would be the complementary angle to 28 degrees, so 62 degrees

Yes, so that is the angle to put in the cosine in your relevant equation.

In your previous thread, a mass went up a 35 degree slope, so in that case the angle between the displacement and gravity was 125 degrees. Taking the cosine produced a negative value, which is correct: gravity did negative work.