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2 years ago

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You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle

30∘ above the horizontal with a force of magnitude 240 N, causing the box to move horizontally. The force of friction between the moving box and the floor has magnitude 41.5 N . What is the box's acceleration just after it begins to move?

1 answer:

charle [14.2K]2 years ago

6 0

Answer:

a= $5.54m/s^{2}$

Explanation:

The net force, $F_{net}$ of the box is expressed as a product of acceleration and mass hence

$F_{net}=ma$ where m is mass and a is acceleration

Making a the subject, a= $\frac {F_{net}}{m}$

From the attached sketch,

∑ $F_{net}=Fcos\theta-F_{f}$ where $F_{f}$ is frictional force and $\theta$ is horizontal angle

Substituting ∑ $F_{net}$ as $F_{net}$ in the equation where we made a the subject

a= $\frac {Fcos\theta-F_{f}}{m}$

Since we’re given the value of F as 240N, $F_{f}$ as 41.5N, $\theta$ as $30^{o}$ and mass m as 30kg

a= $\frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}$

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This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

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