A force $F$ is used to raise a $4kg$ mass $M$ from the ground to a height of $5 m$. What is the work done by the force $F$?

The answer is 200 J, but I got 174 J. I know that Work is $Fd\cos (\theta)$, but I don't understand why in this problem, we are not using $\sin60^\circ$ and only using $Fd\cos(\theta)$. I thought work is calculated based on the force done parallel to the object's movement, which is going up. Could someone please help explain to me where my logic is going wrong? Thank you!!

Here is a picture of the problem:

• The angle is a distractor. All you need to worry about is the value of force $F$ and the displacement of the force which is in the same direction as the force. – Farcher Jul 29 '17 at 19:55
• $2Fcos(\theta)=mg$ is the force that doing the work. So work done is $2Fcos(\theta)*h=mgh$ – Icchyamoy Jul 29 '17 at 20:06
• How do you know that a huge value of $F$ doesn't make the mass pass $4$ metres height at some huge velocity and thus huge kinetic energy? – DJohnM Jul 30 '17 at 1:21
• Welcome to Physics! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – Kyle Kanos Jul 30 '17 at 13:50

I thought work is calculated based on the force done parallel to the object's movement, which is going up

That's true and that's exactly what's happening here: The force being applied to the mass points parallel to the object's movement, independently of the angle given in this example. If you need another anology, maybe this helps: Imagine some weight lifting machine in the gym: It really doesn't matter, at which angle you pull those ropes, you will always do the same work when pulling then, no matter at which angle the ropes are. And that's because- like stated before- the role does deflect the rope, but not the force being applied to the mass.

• If somebody has voted the answer as "not useful", please correct it or give a reasoning for that, because it did answer the above question! – Alienbash Jul 30 '17 at 16:47

The work done depends on the distance moved in the direction of the force, by the point of application of the force. The correct direction to use is the direction at the point of application of the force.

The force is applied at the end of the rope, and this end moves a distance of 5m in the direction of the rope, which is the direction of the force. So the work done by this force is 5F Joules, assuming F is in Newtons. This value is the same whatever angle the rope makes with the horizontal, because the end of the rope always moves in the direction it is pulled.

The rope bends around the pulley and transmits the same force F to the point at which the rope is attached to mass M. The force at this point is applied upwards, and this point moves upwards by 5m. So the work done by force F at this point of application is again 5F Joules.

We are expected to assume that F is the minimum possible force to raise the mass M, which is its weight of about 40N.

Your mistake is that you have used the direction of the force at the free end of the rope, and the point of application where it is attached to mass M. The force has a different direction at these two points.

So yes, the force you pull on the string with is at a downward angle. But the part of the string that is pulling upward on the weight is completely vertical (i.e all of the force is being used to move the weight). So, your theta is zero degrees, and thus $W=Fd\cos(0)=4 kg \cdot 9.8 m/s^2 \cdot 5 m = 196 J$, or $200 J$ if you round it to one sig fig.