Question1. A football player pushed a 75 kg blocking sled with a constant horizontal force of 400 N . The coefficient of kinetic friction between the sled and ground is 0.3 . How much horizontal force opposes the forward motion of the sled? What is the sled's horizontal acceleration? (Assume that the playing surface is level).

Studdy Solution

STEP 1

1. The mass of the blocking sled is 75 kg.
2. The constant horizontal force applied by the football player is 400 N.
3. The coefficient of kinetic friction between the sled and the ground is 0.3.
4. The playing surface is level.
5. We need to find the horizontal force opposing the sled's motion and the sled's horizontal acceleration.

STEP 2

1. Calculate the force of kinetic friction opposing the sled's motion.
2. Determine the net force acting on the sled.
3. Calculate the sled's horizontal acceleration using Newton's second law.

STEP 3

Calculate the force of kinetic friction opposing the sled's motion.
The force of kinetic friction ( $f_k$ ) can be calculated using the formula:
$f_k = \mu_k \times N$
where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force. Since the surface is level, the normal force $N$ is equal to the gravitational force on the sled, which is $m \times g$ , where $m = 75 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$ .
$N = 75 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 735 \, \text{N}$
Now, calculate $f_k$ :
$f_k = 0.3 \times 735 \, \text{N} = 220.5 \, \text{N}$
The horizontal force opposing the forward motion of the sled is $220.5 \, \text{N}$ .

STEP 4

Determine the net force acting on the sled.
The net force ( $F_{\text{net}}$ ) is the difference between the applied force and the force of kinetic friction:
$F_{\text{net}} = F_{\text{applied}} - f_k$ $F_{\text{net}} = 400 \, \text{N} - 220.5 \, \text{N} = 179.5 \, \text{N}$

STEP 5

Calculate the sled's horizontal acceleration using Newton's second law.
Newton's second law states:
$F_{\text{net}} = m \times a$
where $a$ is the acceleration. Rearrange the equation to solve for $a$ :
$a = \frac{F_{\text{net}}}{m}$ $a = \frac{179.5 \, \text{N}}{75 \, \text{kg}}$ $a = 2.3933 \, \text{m/s}^2$
The sled's horizontal acceleration is approximately $2.39 \, \text{m/s}^2$ .

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