Question

How do you find the force needed to accelerate a 0.007 kg pellet from rest to 125 m/s over a distance of 0.8 m along the barrel?

Answer 1

Hey, there!

Hers the answer you are looking for.

From the third equation of motion,

v² - u² = 2as

This implies,

$a = \frac{ {v}^{2} - {u}^{2} }{2s}$

where,

v = final velocity

u = initial velocity

s = displacement

and a = acceleration

Also, Force = mass × acceleration

i.e., F = m × a

$F = m \times \frac{ {v}^{2} - {u}^{2} }{2s}$

Now you just need to put the values in the final equation to get the answer.

$F = \frac{7}{1000} \times \frac{ {125}^{2} - {0}^{2} }{2 \times 0.8} \\ \\ F = \frac{7}{1000} \times \frac{15625}{1.6}$

F = 68.36N (approx.)

Hope this helps.

Thanks.

Answer 2

Answer:

Hers the answer you are looking for.

From the third equation of motion,

v² - u² = 2as

This implies,

a = \frac{ {v}^{2} - {u}^{2} }{2s}a=

2s

v

2

−u

2

where,

v = final velocity

u = initial velocity

s = displacement

and a = acceleration

Also, Force = mass × acceleration

i.e., F = m × a

F = m \times \frac{ {v}^{2} - {u}^{2} }{2s}F=m×

2s

v

2

−u

2

Now you just need to put the values in the final equation to get the answer.

\begin{gathered}F = \frac{7}{1000} \times \frac{ {125}^{2} - {0}^{2} }{2 \times 0.8} \\ \\ F = \frac{7}{1000} \times \frac{15625}{1.6} \end{gathered}

F=

1000

7

×

2×0.8

125

2

−0

2

F=

1000

7

×

1.6

15625

F = 68.36N (approx.)

Hope this helps.

Thanks.