Force is applied to a body of mass $3 \text{ kg}$ at rest on a frictionless horizontal surface as shown in the force against time (F-t) graph. The speed of the body after $1 \text{ s}$ is
Step 1: Understanding the Concept:
The area under a Force-Time graph equals the Impulse delivered to the body.
By the Impulse-Momentum Theorem, impulse equals the change in momentum of the body. Step 2: Key Formula or Approach:
Impulse $J = \int F dt = \text{Area under F-t graph}$.
Change in momentum $\Delta p = mv_f - mv_i = J$.
Since the body starts from rest, $v_i = 0$, so $mv_f = J$. Step 3: Detailed Explanation:
Calculate the area under the graph from $t=0$ to $t=1.0 \text{ s}$. The area is composed of two rectangles.
Area 1 ($t=0$ to $t=0.5$): $A_1 = \text{height} \times \text{width} = 8 \text{ N} \times 0.5 \text{ s} = 4 \text{ N}\cdot\text{s}$.
Area 2 ($t=0.5$ to $t=1.0$): $A_2 = \text{height} \times \text{width} = 4 \text{ N} \times 0.5 \text{ s} = 2 \text{ N}\cdot\text{s}$.
Total Impulse $J = A_1 + A_2 = 4 + 2 = 6 \text{ N}\cdot\text{s}$.
Now, apply the impulse-momentum theorem:
\[ J = \Delta p = m(v_f - v_i) \]
Given mass $m = 3 \text{ kg}$ and initial velocity $v_i = 0$.
\[ 6 = 3(v_f - 0) \]
\[ 6 = 3v_f \]
\[ v_f = \frac{6}{3} = 2 \text{ m/s} \]
Step 4: Final Answer:
The speed is 2 m/s.
