Question

A 4.60-kg sled is pulled across a smooth ice surface. The force acting on the sled is of magnitude 6.20 N and points in a direction 35.0o above the horizontal. If the sled starts at rest, how fast is it going after being pulled for 1.15s?

Answer 1

Answer:

1.27 m/s

Explanation:

The force acting on the sled is F = 6.20 N. Since it is acting at an angle of 35° to the horizontal, its horizontal component which moves the sled forward is Fcos35. This force is the net force on the sled. So,

Fcos35 = ma where m = mass of sled = 4.60 kg and a = acceleration of sled

a = Fcos35/m = 6.20cos35/4.60 = 1.1 m/s².

We now know its acceleration. To find the sleds speed after time t = 1.15 s, we use v = u + at where u = initial velocity of sled = 0 m/s (since it starts from rest)

Substituting the remaining values of a and t into the equation, we have

v = u + at = 0 + 1.1 × 1.15 = 1.27 m/s

So, the sled goes 1.27 m/s fast after 1.15 s

Answer 2

If the sled starts at rest, it is going with a speed of 1.27m/s after being pulled for 1.15secs

In order to get the velocity of the sled after being pulled, we will use the equation of motion

v = u + at

u is the initial velocity

a is the acceleration of the sled

t is the time taken

First, we need to get the acceleration of the sled using Newton's second law;

$\sum F_x = ma_x\\Fcos35^0=4.60a_x\\6.20cos35^0=4.60a_x\\5.0787 = 4.60a_x\\a_x=\frac{5.0787}{4.60}\\a_x= 1.10m/s^2$

Substitute the acceleration into the equation of motion above:

$v = u + at\\v =0+1.1(1.15)\\v=1.27m/s$

Hence if the sled starts at rest, it is going with a speed of 1.27m/s after being pulled for 1.15secs

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