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damaskus [11]
3 years ago
7

A large box has a mass of 500kg and the coefficient of static friction for the box and the floor is 0.45, and the coefficient of

kinetic friction is 0.30.
a. What is the minimum horizontal force needed to get the box moving?
b. If you continue to push with that force, what will the acceleration of the box be?
Please help asap. Thank you.
Physics
1 answer:
Mumz [18]3 years ago
4 0
A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.

b) The acceleration will be the horizontal force - the kinetic friction force (since they act in opposite directions) divided by the mass. Kinetic friction force = (coefficient of kinetic friction)*(normal force or weight). 

F(net) = (2207.25 N)-(0.30)(4905 N) = 735.75 N

a = (735.75 N)/(500kg)= 1.4715 m/s^2
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(a) The height of the cliff will be 8.26 meters.

(b) The time would it take to reach the ground will be 0.717 sec.

<h3>What is velocity?</h3><h3 />

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component.

(a) The height of the cliff will be 8.26 meters.

According to Newton's second equation of motion

\rm H =ut-\frac{1}{2} gt^2 \\\\ \rm H =8\times 2.35-\frac{1}{2} 9.81 (2.35)^2\\\\\rm H =8.16 \; m

Hence The height of the cliff will be 8.26 meters.

(b)The time would it take to reach the ground will be 0.717 sec.

We must have the final velocity to find the time so;

\rm v^2=u^2+2gh\\\\ \rm v^2=8^2+2\times 9.81 \times 8.6 \\\\ \rm v= \sqrt{8^2+2\times 9.81 \times 8.6}\\\\\rm v=15.03 \;m/sec

According to Newton's third equation of motion ;

\rm v=u-gt \\\\ \rm t=\frac{v-u}{g} \\\\ \rm t=\frac{15.03-8}{9.81} \\\\ \rm t=0.717 sec.

Hence the time would it take to reach the ground will be 0.717 sec.

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A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the t
oee [108]

Answer:

v_x = 1.26 m/s

Explanation:

given,

weight of swimmer = 510 N

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speed of the swimmer = ?

horizontal velocity  of the swimmer should be that much it can cross the wedge.

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now,time taken by the swimmer to cover 9 m

initial vertical velocity of the swimmer is zero.

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s = ut +\dfrac{1}{2}gt^2

9= 0+\dfrac{1}{2}\times 9.8\times t^2

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same time will be taken to cover horizontal distance.

now, from equation 1

1.75 = v_x × 1.39

v_x = 1.26 m/s

horizontal speed of the swimmer is equal to 1.26 m/s

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