Force: Mathematical problem & solution-2

Problem 1: Force and Friction

A 10 kg box is being pushed across a horizontal surface. The coefficient of kinetic friction between the box and the surface is 0.3. Calculate the force required to keep the box moving at a constant velocity.

Solution:

1. Calculate the Normal Force:

The normal force (${𝐹}_{𝑁}$) is the force exerted by a surface to support the weight of an object resting on it. For a horizontal surface, it is equal to the weight of the object.

$${𝐹}_{𝑁}=𝑚\cdot 𝑔$$

Given:

• $𝑚=10$ kg
• $𝑔=9.8{\text{ m/s}}^2$

$${𝐹}_{𝑁}=10\text{ kg}\times 9.8{\text{ m/s}}^2=98\text{ N}$$

2. Calculate the Frictional Force:

The frictional force (${𝐹}_{𝑓}$) is given by:

$${𝐹}_{𝑓}=𝜇\cdot {𝐹}_{𝑁}$$

Given:

• $𝜇=0.3$
• ${𝐹}_{𝑁}=98\text{ N}$

$${𝐹}_{𝑓}=0.3\times 98\text{ N}=29.4\text{ N}$$

3. Calculate the Force Required to Keep the Box Moving:

To keep the box moving at a constant velocity, the applied force ($𝐹$) must balance the frictional force.

$$𝐹={𝐹}_{𝑓}=29.4\text{ N}$$

So, the force required to keep the box moving at a constant velocity is $29.4\text{ N}$.

Problem 2: Force and Circular Motion

A 5 kg object is attached to a string and is whirled in a horizontal circle of radius 2 meters at a constant speed of 4 m/s. Calculate the tension in the string.

Solution:

1. Calculate the Centripetal Force:

The centripetal force (${𝐹}_{𝑐}$) required to keep an object moving in a circle is given by:

$${𝐹}_{𝑐}=\frac{𝑚\cdot {𝑣}^2}{𝑟}$$

Given:

• $𝑚=5$ kg
• $𝑣=4\text{ m/s}$
• $𝑟=2$ m

$${𝐹}_{𝑐}=\frac{5\text{ kg}\times (4\text{ m/s}{)}^2}{2\text{ m}}=\frac{5\times 16}{2}=40\text{ N}$$

So, the tension in the string, which provides the centripetal force, is $40\text{ N}$.

Problem 3: Force and Hooke's Law

A spring has a spring constant ($𝑘$) of $500\text{ N/m}$. If the spring is compressed by 0.1 meters, calculate the force exerted by the spring.

Solution:

1. Apply Hooke's Law:

Hooke's Law states that the force exerted by a spring ($𝐹$) is proportional to the displacement ($𝑥$) from its equilibrium position:

$$𝐹=-𝑘\cdot 𝑥$$

Given:

• $𝑘=500\text{ N/m}$
• $𝑥=0.1$ m

$$𝐹=-500\text{ N/m}\times 0.1\text{ m}=-50\text{ N}$$

The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. So, the magnitude of the force is $50\text{ N}$.

Problem 4: Force and Momentum Change

A 2 kg ball moving with a velocity of 10 m/s is brought to rest in 0.5 seconds by a constant force. Calculate the force exerted on the ball.

Solution:

1. Calculate the Change in Momentum:

Momentum ($𝑝$) is given by $𝑝=𝑚\cdot 𝑣$. The change in momentum ($\mathrm{\Delta }𝑝$) is:

$$\mathrm{\Delta }𝑝=𝑚\cdot ({𝑣}_{𝑓}-{𝑣}_{𝑖})$$

Given:

• $𝑚=2$ kg
• Initial velocity, ${𝑣}_{𝑖}=10\text{ m/s}$
• Final velocity, ${𝑣}_{𝑓}=0\text{ m/s}$

$$\mathrm{\Delta }𝑝=2\text{ kg}\times (0-10\text{ m/s})=-20\text{ kg}\cdot \text{m/s}$$

2. Calculate the Force:

Force ($𝐹$) is the rate of change of momentum. Using the formula $𝐹=\frac{\mathrm{\Delta }𝑝}{\mathrm{\Delta }𝑡}$:

$$𝐹=\frac{-20\text{ kg}\cdot \text{m/s}}{0.5\text{ s}}=-40\text{ N}$$

The negative sign indicates that the force is in the direction opposite to the ball's initial motion. So, the magnitude of the force is $40\text{ N}$.

Summary of Examples:

1. Force required to overcome friction: $29.4\text{ N}$
2. Tension in a string during circular motion: $40\text{ N}$
3. Force exerted by a compressed spring: $50\text{ N}$
4. Force to bring a ball to rest: $40\text{ N}$