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A block of mass 1.95 kg slides with an initial speed of 4.33m/s on a smooth horizontal surface. The block now encounters a rough patch with a coefficient of friction of 0.26. What is the acceleration of the block when it is in the rough patch?

User Ahmedhosny
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Final answer:

The acceleration of a 1.95 kg block on a rough patch with a coefficient of friction of 0.26 is approximately 2.5617 m/s² in the opposite direction to its motion. This calculation was derived using force and motion equations considering the force of friction opposing the block's initial velocity.

Step-by-step explanation:

To find the acceleration of the block when it is on the rough patch, we need to use the formula F = ma, where F is the force acting on the block, m is the mass of the block, and a is the acceleration. The only horizontal force acting on the block while it is on the rough patch is the frictional force, which can be calculated using Ffriction = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the block since the surface is horizontal, so N = mg. The frictional force opposing the motion of the block is therefore Ffriction = μ mg.

Given that the mass m = 1.95 kg, the initial speed is not necessary for calculating the acceleration due to friction. The coefficient of friction μ = 0.26, and the acceleration due to gravity g = 9.81 m/s2, the frictional force is:

• Ffriction = (0.26)(1.95 kg)(9.81 m/s2)

• Ffriction = 4.9983 N (approximately)

Now, using F = ma, we isolate a:

• a = Ffriction/m

• a = 4.9983 N / 1.95 kg

• a = 2.5617 m/s2 (approximately)

Thus, the acceleration of the block on the rough patch is about 2.5617 m/s2 in the opposite direction of motion. The negative sign indicates deceleration.

User Antonio Carito
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