Question

A block of mass 5 kg starting from rest pulled up on a smooth incline plane making an angle of 30^◦ with horizontal with an affective acceleration of 1 ms⁻². The power delivered by the pulling force at t = 10 s from the starts is W. [use g=10 ms⁻²] (Calculate the nearest integer value)

Answer 1

Given:

• Mass (\( m \)) = \( 5 \, \text{kg} \)
• Angle of incline (\( \theta \)) = \( 30^\circ \)
• Acceleration (\( a \)) = \( 1 \, \text{m/s}^2 \)
• Time (\( t \)) = \( 10 \, \text{s} \)
• Gravitational acceleration (\( g \)) = \( 10 \, \text{m/s}^2 \)
• Initial velocity (\( u \)) = \( 0 \, \text{m/s} \)
  

Step 1: Resolve Forces and Find Pulling Force (\( F \))

The forces acting along the incline include:

• Pulling force (\( F \)) acting upwards along the incline.
• The component of weight (\( mg \sin \theta \)) acting downwards along the incline.
• The force causing acceleration (\( ma \)) acting upwards.

Applying Newton’s second law along the incline:

\[ F - m g \sin \theta = m a. \]

Substitute the given values (\( m = 5 \, \text{kg}, g = 10 \, \text{m/s}^2, \sin 30^\circ = 0.5, a = 1 \, \text{m/s}^2 \)):

\[ F - 5 \cdot 10 \cdot 0.5 = 5 \cdot 1. \]

Simplify:

\[ F - 25 = 5 \implies F = 30 \, \text{N}. \]

Step 2: Calculate the Final Velocity (\( v \))

Using the first equation of motion:

\[ v = u + a t. \]

Substitute \( u = 0, a = 1 \, \text{m/s}^2, t = 10 \, \text{s} \):

\[ v = 0 + 1 \cdot 10 = 10 \, \text{m/s}. \]

Step 3: Calculate the Power (\( P \))

Power is the rate at which work is done, given by:

\[ P = F v. \]

Substitute \( F = 30 \, \text{N} \) and \( v = 10 \, \text{m/s} \):

\[ P = 30 \cdot 10 = 300 \, \text{W}. \]

Final Answer:

The power delivered by the pulling force is \( 300 \, \text{W} \).