A stationary 1.67-kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt2, where t is th

Question

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A stationary 1.67-kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt2, where t is th

e time in milliseconds from the instant the stick first contacts the object.
If :

a = 1500 N/(ms)

b = 20 N/(ms)2

what is the speed of the object just after it comes away from the stick at t = 2.74 ms?

Physics

1 answer:

serg [1.1K] · Answer 1 · 2026-02-22T12:31:07+03:00

Answer:

$v_{f}$ = 3289.8 m/s

Explanation:

This problem can be approached using momentum definitions.

I = ∫ F dt

We substitute and compute.

I = ∫ (at - bt²) dt

Integrating gives us:

I = a t² / 2 - b t³ / 3

We will evaluate between the limits I=0 for t = 0 ms and higher I=I for t = 2.74 ms:

I = a (2.74² / 2- 0) - b (2.74³ / 3 -0)

I = a 3.754 - b 6.857

Substituting the values for a and b, we find:

I = 1500 3.754 - 20 6.857

I = 5,631 - 137.14

I = 5493.9 N s

Next, we engage the relationship between impulse and momentum:

I = Δp = m $v_{f}$ - m v₀o

I = m $v_{f}$ - 0

$v_{f}$ = I / m

$v_{f}$ = 5493.9 /1.67

$v_{f}$ = 3289.8 m/s