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nordsb [41]
3 years ago
8

A crate pushed along the floor with velocity vâ i slides a distance d after the pushing force is removed. if the mass of the cra

te is doubled but the initial velocity is not changed, what distance does the crate slide before stopping?
Physics
1 answer:
Alenkinab [10]3 years ago
7 0
<span>d The mass is doubled which means that both the momentum and kinetic energy are also doubled. Also the normal force that's acting along with the coefficient of kinetic friction is also doubled. So the friction that's working to slow down the crate is doubled. So the crate will have double the kinetic energy that needs to be dissipated, but the rate of dissipation is also doubled, so the total time required to dissipate the kinetic energy is the same. And since both crates start out with the same velocity and since they'll lose energy (and velocity) at the same proportional rate, they'll take the same distance to slide to a stop.</span>
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A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?
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1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.
harkovskaia [24]

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, F_y = Fsin \theta = 250sin45 = 176.78 \ N

Applied force in x-direction, F_x = Fcos \theta = 250cos45 = 176.78 \ N

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N

Apply Newton's second law of  motion;

F = ma

a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2

(b) the velocity of the crate after 5.0 s

F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s

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