210J
PE is mgh in this context.
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

So now we have an equation and unkown value.
for the thrown rock

for the dropped rock

solving both equation with the quadratic formula:

we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
Answer:
Explanation:
Give that,
Spring constant (k)=40N/m
Force applied =75N
Since the force is applied to the right, we don't know if it is compressing or stretching the spring
So let assume it compress
Using hooke's law
F=-ke
e=-F/k
Then, e=-75/40
e=-1.875m
The deformation is 1.875m.
Let assume it stretch
Using hooke's law
-F=-ke
e=F/k
Then, e=75/40
e=1.875m
The elongation is 1.875m
Assuming motion is on a straight path, the result of two positive components of a vector would also be a positive value since both are having positive signs and directions. The direction would be the same with the motion as well. Hope this answers the question. Have a nice day.