The magnitude of the net force exerted in the x direction on a 3.30 Kg particle

Question

The magnitude of the net force exerted in the x direction on a 3.30 Kg particle varies in time as shown in the figure below (a) Find the impulse imparted to the particle from this force. (b) Find the final velocity the particle attains if it is originally at rest. (c) Find its final velocity is -2.9 m/s (d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s. The force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on the ground as a function of time. Starting form rest, a 57.0 Kg athlete jumps down onto the platform from a height of 0.740 m. While she is in contact with the platform during the time interval 0

Explanation / Answer

(a)
Impulse, = F* t
Area under the Force - time Graph is equal to the impulse.

I = 1/2 * 2 * 4 + 4* 1 + 1/2 * 4 * 2 Ns
I = 12 Ns

(b)
If initial Velocity = 0
Impulse, F* t = m* v
12 = 3.30 * v
v = 3.64
vf - vi = 3.64 m/s
vf = 3.64 m/s

(c)
If vi = -2.9 m/s
vf - vi = 3.64 m/s
vf = 3.64 + 2.9 m/s
vf = 6.54 m/s

(d)
Favg = 12/5
Favg = 2.4 N

please post seperate questions in seperate post only.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.