A particle with a mass of 2.70 kg is acted on by a force Fx acting in the x-direction. If the magnitude of the force varies in time as shown in the figure below, determine the following. WebAssign Plot (a) impulse of the force (in kg · m/s ) kg · m/s (b) final velocity of the particle (in m/s) if it is initially at rest m/s (c) Find the final velocity of the particle (in m/s) if it is initially moving along the x-axis with a velocity of −2.90 m/s. m/s
Not going to be able to give you a precise answer without the plot or formula of force over time.
However, (a) the impulse will equal the area under the curve or the integral of the force over time p = ∫Fdt
Edit due to updated information. An impulse has dimensions of a force acting over a time. The given chart has and area under the curve of ½(2 - 0)(8) + (3 - 2)(8) + ½(5 - 3)(8) = 24 N•s impulse (which also equals 24 kg•m/s, the units of momentum). As the entire curve is above the x axis, the force is positive at all times and the mass accelerates in the positive x direction.
(b) the impulse will equal a change in momentum p = mΔv where p = impulse found in part (a) p = m(vf - vi) where vi = 0 vf = p/m
edit
vf = 24 / 2.70 vf = 8.888... vf = 8.89 m/s
(c) similar to part (b), but vi = -2.90 p = m(vf - (-2.90)) p = m(vf + 2.90) vf = p/m - 2.90
edit
vf = 8.89 - 2.90 vf = 5.99 m/s
I hope this helps. Please remember to vote a Best Answer from among your results.
According to the graph, the force increases from 0 N to 8 N in the first two seconds. During the next second, the force is 8 N. During the last two seconds, the force decreases from 8 N to 0 N. Let’s use the following equation to determine the particle’s velocity at two seconds.
Average force * time = mass * change of velocity 4 * 2 = 2.70 * ∆ v
∆ v = 8 ÷ 2.70
This is approximately 2.96 m/s. Since the initial velocity is 0 m/s, this is the velocity at two seconds. For the next second, the force is 8 N. Let’s use the same equation to determine the increase of its velocity
8 * 1 = 2.70 * ∆ v
∆ v = 8 ÷ 2.70
At 3 seconds, v = 16 ÷ 2.70
This is approximately 5.93 m/s. During the last seconds, the force decreases from 8 N to 0 N
4 * 2 = 2.70 * ∆ v ∆ v = 8 ÷ 2.70
Final velocity = 3 * 8 ÷ 2.70 = 8 8/9th m/s
If the initial velocity is -2.90 m/s, we need to subtract from the answer above.
vf = 24 ÷ 2.70 – 2.90
This is approximately 6 m/s.
The product of force times time is the area of the trapezoid.
Area = ½ * (0 + 8) * 5 = 24
This is the increase of the momentum of the particle. So, this is the impulse. If we divide this number by the mass of the particle, the answer is it final velocity. This is good way to check the answer for this problem. I hope this is helpful for you.