My physics lovers....here you go?

Question

How high will a 1.95 kg rock go if thrown straight up by someone who does 70.0 J of work on it? Neglect air resistance.

Answer should be in meters. I'm really just looking for the equation to solve this with. But answers are encouraged. I can compare my results with yours. Thanks.

Answer 1

So at the highest point, the rock would have the energy of 70 joules.

mgh = 70 joules

1.95kg * 9.8m/sec^2 * h = 70 joules

h = 3.66m

Did you get that also??

Answer 2

Sneaky question! I had to think about this...

If you do 70.0j of work throwing a rock straight up, then neglecting air resistance, then at the highest point, the rock has the whole 70j of energy as Potential Energy (PE) to fall back down again.

PE = m x g x h

You want to know height, so-rearrange the equation to give:

h = PE / m x g (assuming you take g as 10 m/s^2, but sometimes it is taken as 9.8, or even 9.81, although the Internationally agreed standard for g is 9.80665 m/s^2)

= 70 / 1.95 x 10
= 3.59m (to three significant figures)

I hope this helps!

Answer 3

The work done on the 1.95 kg mass will translate into potential energy; the potential energy is given by PE = m * g * h. So you get
70.0 J = m * g * h = (1.95 kg) * (9.8 m/s^2) * h

h = (70.0J)/[1.95 kg * 9.8 m/s^2]

Everything is in SI units, so the answer comes out to be in meters.

Answer 4

W = Fx costheta
theta is 0, so costheta = 1

W = Fx
W = Max
70.0J/Ma = x
x = 3.66 m

Once it reaches its max height, it only has potential energy. Since no energy is being lost, it should still have 70.0J left.