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Nadya [2.5K]
3 years ago
8

Johnson is dragging a bag on a ice surface. He pulls on the strap with a force of 112 N at an angle of 45° to the horizontal to

displace it 84 m in 3.33 minutes. Determine the work done by Johnson on the bag and the power generated by Johnson
Physics
1 answer:
Nady [450]3 years ago
5 0

Answer with Explanation:

We are given that

Force=F=112 N

\theta=45^{\circ}

Distance,s=84 m

Time, t=3.33 minutes

We have to find the work done by Johnson on the bag and the power generated by Johnson.

Work done, W=Fscos\theta

Using the formula

Work done, W=112\times 84cos45=6652.46 J

Power, P=\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt

Using 1 minute=60 s

Hence, the power generated by Johnson=33.3 watt

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Answer:

1034.78 m

Explanation:

The shortest distance is the displacement of the car from initial position to final position.

Displacement of body is given using Newton's equations of motion.

Given:

Initial velocity, u = 120 m/s

Final velocity, v = 0 m/s( As the car stops in its final position)

Coefficient of static friction, \mu=0.71

Acceleration due to gravity, g=9.8\textrm{ } m/s^{2}

Now, when brakes are applied, only friction acts on the body in a direction opposite to that of its motion.

The acceleration of the car when friction acts the stopping force is given as:

a=- \mu g.

The acceleration is negative as it reduces the velocity of the motion and acts in the direction opposite to that of the motion.

Plug in 0.71 for \mu and 9.8 m/s² for g. Solve for a.

So, a=-0.71\times 9.8=-6.958\textrm{ }m/s^{2}.

Now, displacement of the car is given using the following equation of motion:

v^{2}=u^{2} +2aS

Here, S is the displacement of the racing car.

Plug in 120 m/s for u, 0 m/s for v, -6.958 m/s² for a. Solve for S. This gives,

0^{2} =(120)^{2}+2(-6.958)S\\13.916S=14400\\S=\frac{14400}{13.916}=1034.78\textrm{ m}

Therefore, the shortest distance in which she can brake and stop is 1034.78 m.

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We have

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