Math Problem Statement

The force-time graph for the motion of a body of mass 8 kg moving along a straight line is shown in the figure. Find the change in velocity of the body in SI units.

Solution

The question provides a force-time graph for a body with a mass of 8kg8 \, \text{kg} and asks for the change in velocity of the body. To solve this, we will use the following key concepts:

  1. Impulse-Momentum Theorem:
    Impulse (JJ) is the area under the force-time graph, and it is related to the change in momentum (Δp\Delta p) of the body by: J=Δp=mΔvJ = \Delta p = m \Delta v where:

    • JJ is the impulse,
    • mm is the mass of the body, and
    • Δv\Delta v is the change in velocity.
  2. Steps:

    • First, calculate the area under the force-time graph to find the impulse.
    • Then, use the relationship between impulse and change in momentum to find the change in velocity.

Breakdown of the graph:

  • From t=0t = 0 to t=10st = 10 \, \text{s}, the force increases linearly from 0N0 \, \text{N} to 10N10 \, \text{N}.
  • From t=10st = 10 \, \text{s} to t=20st = 20 \, \text{s}, the force is constant at 2N-2 \, \text{N}.

Now, let's calculate the area under each section:

  1. From t=0st = 0 \, \text{s} to t=10st = 10 \, \text{s} (a triangle):

    • The area is given by: Area=12×Base×Height=12×10s×10N=50Ns\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10 \, \text{s} \times 10 \, \text{N} = 50 \, \text{Ns}
  2. From t=10st = 10 \, \text{s} to t=20st = 20 \, \text{s} (a rectangle):

    • The area is given by: Area=Base×Height=10s×(2)N=20Ns\text{Area} = \text{Base} \times \text{Height} = 10 \, \text{s} \times (-2) \, \text{N} = -20 \, \text{Ns}

Total Impulse:

J=50Ns+(20)Ns=30NsJ = 50 \, \text{Ns} + (-20) \, \text{Ns} = 30 \, \text{Ns}

Change in velocity:

Using the impulse-momentum theorem: J=mΔvJ = m \Delta v 30Ns=8kg×Δv30 \, \text{Ns} = 8 \, \text{kg} \times \Delta v Δv=308=3.75m/s\Delta v = \frac{30}{8} = 3.75 \, \text{m/s}

Thus, the change in velocity of the body is 3.75 m/s.


Would you like further details or have any other questions?

Here are five related questions for further practice:

  1. What would be the change in velocity if the mass of the body were 10 kg instead of 8 kg?
  2. How would the result change if the force-time graph had a negative slope after 10 seconds?
  3. How do you calculate impulse from a force that varies non-linearly with time?
  4. How is the impulse-momentum theorem related to Newton's second law of motion?
  5. What happens to the change in velocity if the area under the graph were zero?

Tip: In force-time problems, always focus on calculating the area under the graph to determine the impulse applied to the object.

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Math Problem Analysis

Mathematical Concepts

Impulse-Momentum Theorem
Kinematics

Formulas

J = Δp = m Δv (Impulse-Momentum Theorem)
Impulse = Area under the Force-Time graph
Change in velocity: Δv = J / m

Theorems

Impulse-Momentum Theorem

Suitable Grade Level

Grades 10-12