Question

A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds

Answer 1

Answer:

Part a)

$a = 7.89 m/s^2$

Part b)

$a = 3.51 m/s^2$

Part c)

$\Deltra a = 4.38 m/s^2$

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

$F_f = 5552.8 N$

Explanation:

Part a)

As we know by newton's II law

$F = ma$

here we know that

$m = 1267 kg$

$F = 10,000 N$

Now we have

$a = \frac{F}{m}$

$a = \frac{10,000}{1267}$

$a = 7.89 m/s^2$

Part b)

distance covered by the car

$d = 394.6 m$

t = 15 s

now by kinematics we have

$d = \frac{1}{2}at^2$

$394.6 = \frac{1}{2}a(15^2)$

$a = 3.51 m/s^2$

Part c)

Difference of acceleration is given as

$\Delta a = a_{expected} - a_{real}$

$\Delta a = 7.89 - 3.51$

$\Deltra a = 4.38 m/s^2$

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

Now for magnitude of force is given as

$F - F_f = ma$

$10,000 - F_f = ma$

$10,000 - F_f = 1267\times 3.51$

$F_f = 5552.8 N$

Answer 2

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/$s^{2}$

B. Formula: a=$\frac{V-V_{0} }{t}$ and s=d/t

speed= 394.6/15

s=26.3m/s

a=$\frac{26.3-0}{15}$

a=1.75m/$s^{2}$

C. 7.9-1.75=difference of 6.15m/$s^{2}$

D. The force that most likely caused this difference is friction forces