A block of mass 2 kg is at rest on a horizontal table. The coefficient of friction between the block and the table is 0.1. A horizontal force 3 N is applied to the block. The speed of the block (in m/ s) after it has moved a distance 10 m is _______________.

(Specify the answer upto 3 decimal no.)

Correct answer is '3162'. Can you explain this answer?

Question

A block of mass 2 kg is at rest on a horizontal table. The coefficient of friction between the block and the table is 0.1. A horizontal force 3 N is applied to the block. The speed of the block (in m/ s) after it has moved a distance 10 m is _______________.

(Specify the answer upto 3 decimal no.)

Correct answer is '3162'. Can you explain this answer?

Answer

Problem: Find the speed of a 2 kg block after it has moved a distance of 10 m on a horizontal table with a coefficient of friction of 0.1 and a 3 N applied force.

Solution:

Step 1: Find the force of friction

The force of friction (f) can be found using the formula:

f = coefficient of friction x normal force

where normal force (N) is the force exerted by the table on the block and is equal to the weight of the block (mg).

So, f = 0.1 x 2 kg x 9.81 m/s^2 = 1.962 N

Step 2: Find the net force

The net force (F) on the block is the applied force minus the force of friction:

F = 3 N - 1.962 N = 1.038 N

Step 3: Find the acceleration

The acceleration (a) of the block can be found using Newton's second law:

F = ma

So, a = F/m = 1.038 N / 2 kg = 0.519 m/s^2

Step 4: Find the final velocity

The final velocity (v) of the block can be found using the kinematic equation:

v^2 = u^2 + 2as

where u is the initial velocity (which is zero), s is the distance travelled (10 m) and a is the acceleration found in Step 3.

So, v^2 = 0 + 2 x 0.519 m/s^2 x 10 m = 10.38

v = sqrt(10.38) = 3.225 m/s

Answer: The speed of the block after it has moved a distance of 10 m is 3.225 m/s.

Answer

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