Questions: A particle of mass 3 kg is projected at an initial speed of 10 ms^-1 in the horizontal direction. As it travels, it meets a constant resistance of magnitude 6 N. Calculate the deceleration of the particle and the distance travelled by the time it comes to rest.

Solution Steps

• Use Newton's second law: \( F = ma \)
• Given: \( F = 6 \, \text{N} \) (resistance), \( m = 3 \, \text{kg} \)
• Rearrange to find acceleration \( a \): \( a = \frac{F}{m} \)
• Substitute the values: \( a = \frac{6}{3} = 2 \, \text{m/s}^2 \)
• The deceleration is \( 2 \, \text{m/s}^2 \) (negative because it opposes the motion).

• Use the kinematic equation: \( v = u + at \)
• Given: initial velocity \( u = 10 \, \text{m/s} \), final velocity \( v = 0 \, \text{m/s} \), \( a = -2 \, \text{m/s}^2 \)
• Rearrange to solve for time \( t \): \( t = \frac{v - u}{a} \)
• Substitute the values: \( t = \frac{0 - 10}{-2} = 5 \, \text{s} \)

• Use the kinematic equation: \( s = ut + \frac{1}{2}at^2 \)
• Given: \( u = 10 \, \text{m/s} \), \( a = -2 \, \text{m/s}^2 \), \( t = 5 \, \text{s} \)
• Substitute the values: \( s = 10 \times 5 + \frac{1}{2} \times (-2) \times 5^2 \)
• Simplify: \( s = 50 - 25 = 25 \, \text{m} \)

Final Answer