A 10-kg box starts from rest and is subjected to the horizontal force described in the graph. The coefficients of static and kinetic friction between the box and the surface are both 0.2. Determine how fast the box is moving at 12 s (hint: the block won’t start moving until the force of static friction is overcome).**Graph Description:**- The graph depicts a force (F) applied to the box over time (t).- The horizontal axis represents time, ranging from 0 to 12 seconds.- The vertical axis represents force in Newtons (N).- Initially, from 0 to 4 seconds, the force increases linearly to 50 N.- From 4 to 8 seconds, the force remains constant at 50 N.- From 8 to 12 seconds, the force decreases back to 0 N.**Explanation:**The box will only start moving once the applied force exceeds the maximum static frictional force. Given the mass of the box is 10 kg and the coefficient of static friction is 0.2, calculate the force of static friction as follows:\[ F_{\text{friction}} = \mu_s \times m \times g \]Where:- \( \mu_s = 0.2 \) (coefficient of static friction)- \( m = 10 \, \text{kg} \) (mass of the box)- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)This frictional force must be overcome before the box starts moving. Once in motion, calculate the velocity at 12 s by considering the net force (applied force minus kinetic friction) and applying Newton's second law of motion.

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A 10-kg box starts from rest and is subjected to the horizontal force described in the graph. The coefficients of static and kinetic friction between the box and the surface are both 0.2. Determine how fast the box is moving at 12 s (hint: the block won't start moving until the force of static friction is overcome).

A 10-kg box starts from rest and is subjected to the horizontal force described in the graph. The coefficients of static and kinetic friction between the box and the surface are both 0.2. Determine how fast the box is moving at 12 s (hint: the block won't start moving until the force of static friction is overcome).

Elements Of Electromagnetics

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ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Question

A 10-kg box starts from rest and is subjected to the horizontal force described in the graph. The coefficients of static and kinetic friction between the box and the surface are both 0.2. Determine how fast the box is moving at 12 s (hint: the block won’t start moving until the force of static friction is overcome).

**Graph Description:**

- The graph depicts a force (F) applied to the box over time (t).
- The horizontal axis represents time, ranging from 0 to 12 seconds.
- The vertical axis represents force in Newtons (N).
- Initially, from 0 to 4 seconds, the force increases linearly to 50 N.
- From 4 to 8 seconds, the force remains constant at 50 N.
- From 8 to 12 seconds, the force decreases back to 0 N.

**Explanation:**

The box will only start moving once the applied force exceeds the maximum static frictional force. Given the mass of the box is 10 kg and the coefficient of static friction is 0.2, calculate the force of static friction as follows:

\[ F_{\text{friction}} = \mu_s \times m \times g \]

Where:
- \( \mu_s = 0.2 \) (coefficient of static friction)
- \( m = 10 \, \text{kg} \) (mass of the box)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

This frictional force must be overcome before the box starts moving. Once in motion, calculate the velocity at 12 s by considering the net force (applied force minus kinetic friction) and applying Newton's second law of motion.

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