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A 4 kg block is being dragged across a floor by a constant force of 88 N. The coefficient of kinetic friction is 0.4. Find the acceleration of the block. A. 7 m/s² B. 9 m/s² C. 5 m/s² D. 11 m/s²

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The acceleration of the block is approximately 18.076 m/s².

To find the acceleration of the block being dragged across the floor, we can use Newton's second law of motion, which states:

F = ma

Where:

F = force applied (88 N)

m = mass of the block (4 kg)

a = acceleration

First, we need to account for the frictional force opposing the motion. The frictional force can be calculated using the equation:

Frictional Force (F_friction) = coefficient of kinetic friction (μk) × Normal Force (N)

The normal force (N) is the force exerted by the surface to support the weight of the object and is equal to the weight of the object (mg), where g is the acceleration due to gravity (approximately 9.81 m/s²):

N = mg

N = 4 kg × 9.81 m/s²

N ≈ 39.24 N

Now, calculate the frictional force:

F_friction = μk × N

F_friction = 0.4 × 39.24 N

F_friction ≈ 15.696 N

Now, we can calculate the net force acting on the block:

Net Force (F_net) = Applied Force - Frictional Force

F_net = 88 N - 15.696 N

F_net ≈ 72.304 N

Finally, use Newton's second law to calculate the acceleration:

F_net = ma

72.304 N = 4 kg × a

Now, solve for acceleration (a):

a = 72.304 N / 4 kg

a ≈ 18.076 m/s²

So, the acceleration of the block is approximately 18.076 m/s². None of the provided options (A, B, C, D) match this value.

User Mike Goodwin
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