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1 year ago

An airplane starts from rest and accelerates at a constant rate of 3.0 m/s^2 for 30.0 s before leaving the ground.

Physics

1 answer:

Andreas93 [3]1 year ago

6 0

Answer:

90m/s

Explanation:

acceleration depicts the rate of change in velocity per time, therefore an acceleration of 3m/s² means the object increases its velocity 3m/s every second therefore the second squared in the denominator

acceleration = (3m/s)/1s =3m/s²

Therefore, if the plane starts from rest (initial speed = 0m/s) and it increases 3m/s every second for 30 seconds you would have:

final_speed = initial_speed + 3m/s²*30s =0+3*30 m*s/s² = 90 m/s

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3 years ago

A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric

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Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

$F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)$

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

$0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N$

$F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N$

$W_{F_x} = F_xd = 227.5\times 4.8 = 1092J$

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

$W_g = -mg\sin(30^\circ)d$

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

$W_g = -823J$

The energy dissipated by the frictional force can be found as follows:

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

$0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N$

$W_F = Fd = 227.5\times 4.8 = 1092J$

$W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J$

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

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