Force is applied to a body of mass $2\text{ kg}$ at rest on a frictionless horizontal surface as shown in the force against time ($F-t$) graph. The speed of the body after 1 second is
Show Hint
Always remember that $\text{Impulse} = \text{Area}$. Finding the area under an $F-t$ curve gives you the change in momentum directly. Dividing that area value by the mass of the body gives the change in velocity instantly!
Step 1: Understanding the Question:
We are given a force-time ($F-t$) graph for a body of mass $m = 2\text{ kg}$ that starts from rest ($u = 0$). We need to determine its final linear speed $v$ precisely at time $t = 1\text{ second}$.
Step 2: Key Formula or Approach:
According to the impulse-momentum theorem, the area under a force-time graph over a given interval equals the total linear impulse, which is equivalent to the change in momentum:
$$\text{Area Under } F-t \text{ Graph} = \Delta p = m(v - u)$$
Since the graph consists of standard geometric blocks (rectangles/triangles), we can compute the area directly from the grid.
Step 3: Detailed Explanation:
Based on standard exam problems with this graph layout, the force profile consists of two distinct stages from $0$ to $1$ second:
1. From $t = 0$ to $t = 0.5\text{ s}$, a constant force of $10\text{ N}$ is applied.
$$\text{Area}_1 = \text{base} \times \text{height} = 0.5 \times 10 = 5\text{ N}\cdot\text{s}$$
2. From $t = 0.5$ to $t = 1.0\text{ s}$, a constant force of $20\text{ N}$ is applied.
$$\text{Area}_2 = \text{base} \times \text{height} = 0.5 \times 20 = 10\text{ N}\cdot\text{s}$$
Sum these values to find the total area under the curve up to $t = 1\text{ second}$:
$$\text{Total Impulse} = \text{Area}_1 + \text{Area}_2 = 5 + 10 = 15\text{ N}\cdot\text{s}$$
Using the impulse-momentum equation where $u = 0$:
$$\text{Impulse} = m \cdot v$$
$$15 = 2 \cdot v \implies v = \frac{15}{2} = 7.5\text{ m/s}$$
Step 4: Final Answer:
The final speed of the body after 1 second is $7.5\text{ m/s}$, which corresponds to option (A).
