The magnitude of the net force exerted in the x direction on a 2.10-kg particle varies in time as shown in the figure below. F (N) 4 3 A 2 1 1 2 3 0 4 5 (a) Find the impulse of the force over the 5.00-s time interval. I=12 XN S t (s) (b) Find the final velocity the particle attains if it is originally at rest. V=3.076 X m/s avg (c) Find its final velocity if its original velocity is -2.30 î m/s. Vf -0.024 = X m/s (d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s. F₁ 2.4 = XN
The magnitude of the net force exerted in the x direction on a 2.10-kg particle varies in time as shown in the figure below. F (N) 4 3 A 2 1 1 2 3 0 4 5 (a) Find the impulse of the force over the 5.00-s time interval. I=12 XN S t (s) (b) Find the final velocity the particle attains if it is originally at rest. V=3.076 X m/s avg (c) Find its final velocity if its original velocity is -2.30 î m/s. Vf -0.024 = X m/s (d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s. F₁ 2.4 = XN
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![The magnitude of the net force exerted in the \( x \) direction on a 2.10-kg particle varies in time as shown in the figure below.
**Graph Explanation:**
The graph shows force \( F(N) \) on the vertical axis and time \( t(s) \) on the horizontal axis. The graph is shaped like a trapezoid with the following points:
- At \( t = 0 \), \( F = 0 \) N
- At \( t = 1 \), \( F = 3 \) N
- At \( t = 2 \) and \( t = 3 \), \( F = 4 \) N
- At \( t = 4 \), \( F = 3 \) N
- At \( t = 5 \), \( F = 0 \) N
### Problems:
(a) **Find the impulse of the force over the 5.00-s time interval.**
\[
\vec{I} = 12 \, \text{N} \cdot \text{s}
\]
(b) **Find the final velocity the particle attains if it is originally at rest.**
\[
\vec{v}_f = 3.076 \, \text{m/s}
\]
(c) **Find its final velocity if its original velocity is -2.30 \( \hat{\imath} \) m/s.**
\[
\vec{v}_f = -0.024 \, \text{m/s}
\]
(d) **Find the average force exerted on the particle for the time interval between 0 and 5.00 s.**
\[
\vec{F}_\text{avg} = 2.4 \, \text{N}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4dca08d-d7c1-4511-81d7-e3b17171b463%2Fe95aa542-bb27-4ba7-8ebb-7bc8ee9682c7%2Fpc1m59r_processed.png&w=3840&q=75)
Transcribed Image Text:The magnitude of the net force exerted in the \( x \) direction on a 2.10-kg particle varies in time as shown in the figure below.
**Graph Explanation:**
The graph shows force \( F(N) \) on the vertical axis and time \( t(s) \) on the horizontal axis. The graph is shaped like a trapezoid with the following points:
- At \( t = 0 \), \( F = 0 \) N
- At \( t = 1 \), \( F = 3 \) N
- At \( t = 2 \) and \( t = 3 \), \( F = 4 \) N
- At \( t = 4 \), \( F = 3 \) N
- At \( t = 5 \), \( F = 0 \) N
### Problems:
(a) **Find the impulse of the force over the 5.00-s time interval.**
\[
\vec{I} = 12 \, \text{N} \cdot \text{s}
\]
(b) **Find the final velocity the particle attains if it is originally at rest.**
\[
\vec{v}_f = 3.076 \, \text{m/s}
\]
(c) **Find its final velocity if its original velocity is -2.30 \( \hat{\imath} \) m/s.**
\[
\vec{v}_f = -0.024 \, \text{m/s}
\]
(d) **Find the average force exerted on the particle for the time interval between 0 and 5.00 s.**
\[
\vec{F}_\text{avg} = 2.4 \, \text{N}
\]
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