Question

An average frictional force of 80N is required to stop an object at a distance of 25m. If the initial speed of the object is 20m/s,the mass of the object is:

• A. 25Kg
• B. 12Kg
• C. 30Kg
• D. 40Kg
• E. 10Kg

Answer 1

Answer: (E)

Given:

• Frictional force, \( F = 80 \, \text{N} \)
• Stopping distance, \( d = 25 \, \text{m} \)
• Initial speed, \( v_0 = 20 \, \text{m/s} \)
• Final speed, \( v = 0 \, \text{m/s} \) (object comes to rest)

Step 1: Calculate Deceleration

Using the kinematic equation:

\[ v^2 = v_0^2 + 2ad \]

Substitute known values:

\[ 0 = (20)^2 + 2a(25) \]

\[ 0 = 400 + 50a \]

\[ a = -\frac{400}{50} = -8 \, \text{m/s}^2 \]

(Negative sign indicates deceleration)

Step 2: Relate Force and Mass

Using Newton's Second Law:

\[ F = ma \]

\[ 80 = m \times 8 \]

\[ m = \frac{80}{8} = 10 \, \text{kg} \]

Conclusion:

The mass of the object is 10 kg.

Answer: \(\boxed{E}\)

Answer 2

1. Define variables and given information:

• f = 80 N (frictional force)
• d = 25 m (distance)
• v_i = 20 m/s (initial speed)
• v_f = 0 m/s (final speed, since the object stops)
• m = ? (mass of the object)

2. Use the work-energy theorem:

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by friction is negative (since it opposes motion) and brings the object to rest.

\[W = \Delta KE\]

\[-f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\]

3. Plug in the given values and solve for m:

Since \(v_f = 0\), the equation simplifies to:

\[-f \times d = - \frac{1}{2}mv_i^2\]

\[-(80 \, N)(25 \, m) = -\frac{1}{2}m(20 \, m/s)^2\]

\[-2000 = -\frac{1}{2}m(400)\]

\[2000 = 200m\]

\[m = \frac{2000}{200} = 10 \, kg\]