A sled (mass 480 kg) is pulled across a stone floor with a coefficient of kineti

Question

A sled (mass 480 kg) is pulled across a stone floor with a coefficient of kinetic friction of 1.1. The rope that is used to pull it is at an angle a of 30 degrees with the horizontal. When the sled moves at a constant velocity (i.e.. no acceleration). the normal force that the floor exerts on the sled is ... higher than the sled's weight. equal to the sled's weight. lower than the sled's weight. Now you want to make the sled speed up as indicated. What does the magnitude of the required pulling force not depend on? The current velocity of the sled The coefficient of kinetic friction between the sled and the floor The angle at which you pun The mass of the sled How hard (i.e.. with what magnitude force) do you need to pull to make the sled up speed up with an acceleration of 12.10 m/s^2?

Explanation / Answer

m = 480 kg, uk = 1.1 , theta = 30 degrees

Let us consider F is apllied force

Net force along vertical direction

N +Fsin(30) = mg

Normal force N = mg - Fsin(30)

N< mg

Correct option is lower than the sled's weight

Net force along horizontal direction

-FK +Fcos(30) = ma

-uk*(mg-F*sin(30)) +Fcos(30)= ma

Correct option is current velocity of the sled

a = 12.10 m/s^2

-uk*(mg-F*sin(30)) +Fcos(30)= ma

-1.1*(480*9.8 - F*sin(30)) +Fcos(30) = 480*12.1

F = 7756 N

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