Answer:
The required work done is 
Explanation:
Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '
' be the coefficient of friction, then
where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.
Since the application of force by the movers does not create any acceleration to the block, we can write
So the work done (W) in moving the crate by a distance s = 10.6 m is
Answer:
692.31 N
Explanation:
Applying,
F = ma............... Equation 1
Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player
But,
a = (v-u)/t............ Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
F = m(v-u)/t............ Equation 3
From the question,
Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s
Substitute these values into equation 3
F = 75(0-6)/0.65
F = -692.31 N
Hence the average force required to stop the player is 692.31 N
Potential energy is defined by formula
here
m = mass
g = acceleration due to gravity
h = height
Now here two different stones are located at same height
while mass of stone A is twice that of stone B
so here we can say potential energy of A is
Similarly potential energy of B is
now if we take the ratio of two energy
so we can say potential energy of stone A is two times the potential energy of B
Right, as you mentioned in the comments, you find 
by plugging in the different values of 
.
For 
, we have
Similarly, for 
, you get
