A heavy box of mass 50 kg is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.3, then the force of kinetic friction is:

Question

A charged particle moving with a velocity \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\) in a magnetic field \(\vec{B}\) experiences a force \(\vec{F} = F_1\hat{i} + F_2\hat{j}\). Here \(v_1, v_2, F_1, F_2\) are all constants. Then \(\vec{B}\) can be:

Answer 1

Step 1: Formula for Kinetic Friction Force
The force of kinetic friction \( F_k \) is calculated using the formula: \[ F_k = \mu_k \cdot N \] Here, \( \mu_k \) represents the coefficient of kinetic friction, and \( N \) is the normal force.
Step 2: Calculate the Normal Force
On a horizontal surface, the normal force \( N \) equals the object's weight. This is determined by multiplying the mass \( m \) by the acceleration due to gravity \( g \): \[ N = m \cdot g = 50 \times 9.8 = 490 \, \text{N} \]
Step 3: Calculate the Force of Kinetic Friction
Substitute the known values into the kinetic friction formula: \[ F_k = 0.3 \times 490 = 147 \, \text{N} \]Final Answer: The calculated force of kinetic friction is 147 N.

A heavy box of mass 50 kg is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.3, then the force of kinetic friction is:

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The Correct Option is B

Solution and Explanation

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