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baherus [9]
3 years ago
12

An 1120 kg car traveling at 17.2 m/s is brought to a stop while skidding 40m. Calculate the work done on the car by the friction

forces.
Physics
1 answer:
Nesterboy [21]3 years ago
8 0

Answer:

Work = 165670.4 J = 165.67 KJ

Explanation:

First, we will find the deceleration of the car, using the 3rd equation of motion:

2as = v_{f}^2 - v_{i}^2\\

where,

a = deceleration = ?

s = skid distance = 40 m

vf = final speed = 0 m/s

vi = initial speed = 17.2 m/s

Therefore,

2a(40\ m) = (0\ m/s)^2 - (17.2\ m/s)^2\\a = - 3.698\ m/s^2

the negative sign indicates deceleration here.

Now, we will calculate the braking force applied by the brakes on the car:

F = ma\\F = (1120\ kg)(-3.698\ m/s^2)\\F = - 4141.76\ N

the negative sign indicates braking force.

Now, we will calculate the work done using the magnitude of this force:

Work = |F|s\\Work = (4141.76\ N)(40\ m)\\

<u>Work = 165670.4 J = 165.67 KJ</u>

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A ball is thrown upwards from the edge of a cliff. The horizontal velocity and vertical velocity are both 20m/s the distance fro
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6.9s

Explanation:

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The time it takes for the ball to travels to the top is the time it takes for it to decelerate from 20m/s to 0m/s with gravitational deceleration g = 10m/s2

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d = \frac{gt_2^2}{2}

t^2 = \frac{2d}{g} = \frac{2*120}{10} = 24

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