Check my answers please (3 work questions)

[tony873004]

I think I got these right, but I'd feel better if someone could tell me if I did them correctly:

A projectile is launched from ground level at a 45 degree angle. How much work does gravity do on the projectile between its launch and when it hits the ground?

Gravity will do as much work on the object as the work performed by the
y-component of the force that launched the projectile.

W_{gravity} =\Delta K_{gravity}
\]
\[
\Delta K_{gravity} =\frac{1}{2}mv_y^2
\]
\[
\Delta K_{gravity} =\frac{1}{2}m\left( {\sin (45)v} \right)^2
\]
\[
W_{gravity} =\frac{1}{2}m\left( {\sin (45)v} \right)^2


A hockey player pushes a puck of mass 0.50 kg across the ice using a constant force of 10.0 N over a distance of 0.50 m. How much work does the hockey player do? If the puck was initially stationary, what is its final speed? (Ignore friction.)

W=Fd, W=10N*0.5m, W=5N

F=ma
\quad
\Rightarrow
\quad
a=\frac{F}{m}
\]
\[
a=\frac{10.0N}{0.50kg}
\quad
\Rightarrow
\quad
a=\frac{10.0\rlap{--} {k}\rlap{--} {g}\cdot m/s^2}{0.50\rlap{--}
{k}\rlap{--} {g}}
\]
\[
a=20m/s^2
\]
\[
v_f^2 =v_i^2 +2a\Delta d
\quad
\Rightarrow
\quad
v=\sqrt {v_i^2 +2a\Delta d}
\]
\[
v=\sqrt {\left( {2\cdot 20\frac{m}{s^2}} \right)\cdot 0.50m}
\]
\[
v=4.5m/s


If a constant force F=(30.N)i, + (50.N)j acts on a particle that undergoes a displacement (4.0m)i + (1.0m)j , how much work is done on the particle?


W=Fd
\]
\[
W=\sqrt {\left( {3.0N} \right)^2+\left( {5.0N} \right)^2} \cdot \sqrt
{\left( {4.0m} \right)^2+\left( {1.0m} \right)^2}
\]
\[
W=20J

[whozum]

I think I got these right, but I'd feel better if someone could tell me if I did them correctly:

A projectile is launched from ground level at a 45 degree angle. How much work does gravity do on the projectile between its launch and when it hits the ground?

Gravity will do as much work on the object as the work performed by the
y-component of the force that launched the projectile.

W_{gravity} =\Delta K_{gravity}
\]
\[
\Delta K_{gravity} =\frac{1}{2}mv_y^2
\]
\[
\Delta K_{gravity} =\frac{1}{2}m\left( {\sin (45)v} \right)^2
\]
\[
W_{gravity} =\frac{1}{2}m\left( {\sin (45)v} \right)^2



This is all right.


A hockey player pushes a puck of mass 0.50 kg across the ice using a constant force of 10.0 N over a distance of 0.50 m. How much work does the hockey player do? If the puck was initially stationary, what is its final speed? (Ignore friction.)

W=Fd, W=10N*0.5m, W=5N

F=ma
\quad
\Rightarrow
\quad
a=\frac{F}{m}
\]
\[
a=\frac{10.0N}{0.50kg}
\quad
\Rightarrow
\quad
a=\frac{10.0\rlap{--} {k}\rlap{--} {g}\cdot m/s^2}{0.50\rlap{--}
{k}\rlap{--} {g}}
\]
\[
a=20m/s^2
\]
\[
v_f^2 =v_i^2 +2a\Delta d
\quad
\Rightarrow
\quad
v=\sqrt {v_i^2 +2a\Delta d}
\]
\[
v=\sqrt {\left( {2\cdot 20\frac{m}{s^2}} \right)\cdot 0.50m}
\]
\[
v=4.5m/s


Correct, slight rounding error but thats cool.


If a constant force F=(30.N)i, + (50.N)j acts on a particle that undergoes a displacement (4.0m)i + (1.0m)j , how much work is done on the particle?


W=Fd
\]
\[
W=\sqrt {\left( {3.0N} \right)^2+\left( {5.0N} \right)^2} \cdot \sqrt
{\left( {4.0m} \right)^2+\left( {1.0m} \right)^2}
\]
\[
W=20J


W is actually the dot product between F and D. Dot the two vectors. The dot product of two vectors a and b in two dimensions is:

\vec{a} \bullet \vec{b} = a_xb_x+a_yb_y Also I dont know if its a typo, but in your square root you have magnitudes 5.0N and 3.0N for the force, but in the problem they are 50N and 30N.

[tony873004]

Thanks, whozum. :smile:

That is a typo. 3.0 should be in the question, not 30. 50 too.

Isn't d a displacement, and therefore a scalar, not a vector? So the dot product wouldn't apply and it's just straight forward multiplication?

The rounding error, I actually got 4.4721, but rounded t 4.5 because the inputs were all 2 significant figures. Is that what you meant by rounding error, or did I make another mistake? :yuck:

[whozum]

Thanks, whozum. :smile:

That is a typo. 3.0 should be in the question, not 30. 50 too.

Isn't d a displacement, and therefore a scalar, not a vector? So the dot product wouldn't apply and it's just straight forward multiplication?

I just use D because thats what I did in algebra based physics back in high school, it hasnt left me since. The correct equation for work is:

W = \int_{x_1}^{x_2}{F}\bullet{dx} = \int_{x_1}^{x_2}{Fcos(\theta){dx}
Only the component of force parallel to the path will contribute to the amount of work done. The dot product is necessary.

Given a straight path dx which is always parallel (theta = 0) to the (assumed constant) force:

W = \int_{x_1}^{x_2}{F\bullet{dx} = \int_{x_1}^{x_2}{Fcos(0){dx} = \int_{x_1}^{x_2}{F}{dx} = Fx]_{x_1}^{x_2}

The rounding error, I actually got 4.4721, but rounded t 4.5 because the inputs were all 2 significant figures. Is that what you meant by rounding error, or did I make another mistake? :yuck:
Thats what I meant.

[tony873004]

Thanks a lot whozum. I'll switch that to the dot product.

Our book is picky about significant figures. I've come up with answers like 14.57 m/s, and turn to the back to see the answer given as 10 m/s. Luckily, the homework grader does not care about sig figs.

[Doc Al]

A projectile is launched from ground level at a 45 degree angle. How much work does gravity do on the projectile between its launch and when it hits the ground?

Gravity will do as much work on the object as the work performed by the
y-component of the force that launched the projectile.

W_{gravity} =\Delta K_{gravity}
\]
\[
\Delta K_{gravity} =\frac{1}{2}mv_y^2
\]
\[
\Delta K_{gravity} =\frac{1}{2}m\left( {\sin (45)v} \right)^2
\]
\[
W_{gravity} =\frac{1}{2}m\left( {\sin (45)v} \right)^2

Careful with this one. The work done by gravity (which is the only force acting on the projectile) equals the change in KE. When the the projectile returns to ground, how has its KE changed? So what's the net work done by gravity?
A hockey player pushes a puck of mass 0.50 kg across the ice using a constant force of 10.0 N over a distance of 0.50 m. How much work does the hockey player do? If the puck was initially stationary, what is its final speed? (Ignore friction.)

W=Fd, W=10N*0.5m, W=5N

F=ma
\quad
\Rightarrow
\quad
a=\frac{F}{m}
\]
\[
a=\frac{10.0N}{0.50kg}
\quad
\Rightarrow
\quad
a=\frac{10.0\rlap{--} {k}\rlap{--} {g}\cdot m/s^2}{0.50\rlap{--}
{k}\rlap{--} {g}}
\]
\[
a=20m/s^2
\]
\[
v_f^2 =v_i^2 +2a\Delta d
\quad
\Rightarrow
\quad
v=\sqrt {v_i^2 +2a\Delta d}
\]
\[
v=\sqrt {\left( {2\cdot 20\frac{m}{s^2}} \right)\cdot 0.50m}
\]
\[
v=4.5m/s

OK. Two comments. (1) Work is measured in Joules, not Newtons. (2) Why didn't you use \mbox{Work} = \Delta \mbox{Kinetic Energy} = 1/2 m v^2 to solve for the speed; since you found the work, why not use it?

If a constant force F=(30.N)i, + (50.N)j acts on a particle that undergoes a displacement (4.0m)i + (1.0m)j , how much work is done on the particle?


W=Fd
\]
\[
W=\sqrt {\left( {3.0N} \right)^2+\left( {5.0N} \right)^2} \cdot \sqrt
{\left( {4.0m} \right)^2+\left( {1.0m} \right)^2}
\]
\[
W=20J

This is incorrect, as it assumes that the force and displacement are parallel, which is not the case. Take the dot product of the two vectors. (Yes, displacement is a vector.) Since the force is constant there is no need for any integration.

[tony873004]

Thanks, Doc! That makes sense.