The wall exerts a force of equal magnitude but in the opposite direction. So the force by the wall is 10 N to the right.
Answer:
391.5 J
Explanation:
The amount of work done can be calculated using the formula:
- W = F║d
- where the force is parallel to the displacement
Looking at the formula, we can see that the mass of the object does not affect the work done on it.
Substitute the force applied and the displacement of the object into the equation.
- W = (87 N)(4.5 m)
- W = 391.5 J
The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.
Answer:
statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.
Explanation:
The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.
The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction
here, the downward direction signifies the downward motion parallel to the inclined plane.
Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.
Hence, for the block to stop sliding the the above statement should be true.
(a)
consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.
m = mass of the tennis ball = 60 g = 0.060 kg
v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s
v = final velocity of the tennis ball after being hit by racket = - 39 m/s
ΔP = change in momentum of the ball
change in momentum of the ball is given as
ΔP = m (v - v₀)
inserting the above values
ΔP = (0.060) (- 39 - 20)
ΔP = - 3.54 kgm/s
hence , magnitude of change in momentum : 3.54 kgm/s
