Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
11m/s Bc of the fact that he sees her running at 11m/s
Impulse = Force * times and also Impulse = change in momentum.
Given that the mass does not change, change if momentum = mass * (final velocity - initial velocity)
Given that you know mass and initial velocity (which is the velicity before the cart hits the wall) you need the final velocity (which is the velocity after the cart hits the wall).
Answer: the velocity of the cart after it hits the wall.
Answer:
Part a)
Part b)
Explanation:
Part a)
If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block
so we have
Part b)
If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block
so we have
