Jane pushes a large crate of mass 40 kg along the floor in a straight line. The crate starts from rest and after 6 meters it is moving with a velocity of 0.5 m/s in the direction that Jane is pushing. What is the work that Jane has done on the crate after 6 m?
Here's my attempt:
We know $\Delta K = W =0.5 m(v_{f}^2-v_{i}^2)$ ; Plugging in the numbers give us: $$ 0.5 (40) (0.5^2)- 0.5(40)(0^2)=5 J$$
Where would the 6 meters here go or does it not matter?
Jane then lets go of the crate, and Jack immediately starts pulling the crate with a rope in the same direction along the ground as Jane was going, but at angle of 60 degrees above the horizontal. If Jack pulls with a constant magnitude of 10 N, what is the speed of the crate after Jack has pulled it for an additional 10 m?
Here I just took the integral of 10cos(60) from 0 to 10. that gives us 50 J.
Is my understanding of using work-energy theorem correct? If not, what needs to be modified?
1) Let the constant acceleration be $a.$
Speed : $v = at$ , $t$ is time , initially at rest, I.e. $v_0=0$.
Distance : $s=(1/2)at^2$, start from $s_0=0.$
Eliminate $t: s= (1/2)v^2/a$, or $(1/2)v^2 =sa.$
In your case:
Work done: $(1/2)mv^2= (ma)s= Fs$, where $F=ma$ is the constant force applied to accelerate.
2)Jack: F: Force in the direction of motion:
Work done : $W_{Jack} = Fd$, $d=$distance.
Total energy: $E_t =W_{Jane} + W_{Jack}.$
Finally : $(1/2)mV^2_{final} =E_t.$
$V_{final} = \sqrt{2E_t/m}$