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Math  /  Algebra

Question40. (II) At the instant a race began, a 65−kg65-\mathrm{kg} sprinter exerted a force of 720 N on the starting block at a 22∘22^{\circ} angle with respect to the ground. (a) What was the horizontal acceleration of the sprinter? (b) If the force was exerted for 0.32 s , with what speed did the sprinter leave the starting block?

Studdy Solution

STEP 1

1. The sprinter has a mass of 65 kg.
2. The force exerted by the sprinter is 720 N at a 22∘22^{\circ} angle with respect to the ground.
3. The force is exerted for 0.32 seconds.
4. We are assuming no other forces (like friction or air resistance) are acting on the sprinter horizontally.
5. We need to find the horizontal acceleration and the speed at which the sprinter leaves the starting block.

STEP 2

1. Resolve the force into horizontal and vertical components.
2. Calculate the horizontal acceleration using Newton's second law.
3. Calculate the final speed using the horizontal acceleration and time.

STEP 3

Resolve the force into horizontal and vertical components.
The horizontal component of the force Fhorizontal F_{\text{horizontal}} is given by: Fhorizontal=F⋅cos⁡(θ) F_{\text{horizontal}} = F \cdot \cos(\theta) where F=720 N F = 720 \, \text{N} and θ=22∘ \theta = 22^{\circ} .
Calculate: Fhorizontal=720⋅cos⁡(22∘) F_{\text{horizontal}} = 720 \cdot \cos(22^{\circ})

STEP 4

Calculate the horizontal acceleration using Newton's second law.
Newton's second law states: F=m⋅a F = m \cdot a
So, the horizontal acceleration a a is: a=Fhorizontalm a = \frac{F_{\text{horizontal}}}{m}
Substitute the known values: a=720⋅cos⁡(22∘)65 a = \frac{720 \cdot \cos(22^{\circ})}{65}

STEP 5

Calculate the final speed using the horizontal acceleration and time.
The final speed v v can be calculated using the equation: v=u+a⋅t v = u + a \cdot t
where u=0 m/s u = 0 \, \text{m/s} (initial speed), a a is the horizontal acceleration from Step 2, and t=0.32 s t = 0.32 \, \text{s} .
Calculate: v=0+(720⋅cos⁡(22∘)65)⋅0.32 v = 0 + \left(\frac{720 \cdot \cos(22^{\circ})}{65}\right) \cdot 0.32
Solution:
(a) The horizontal acceleration of the sprinter is: a=720⋅cos⁡(22∘)65≈10.03 m/s2 a = \frac{720 \cdot \cos(22^{\circ})}{65} \approx 10.03 \, \text{m/s}^2
(b) The speed with which the sprinter leaves the starting block is: v≈10.03⋅0.32≈3.21 m/s v \approx 10.03 \cdot 0.32 \approx 3.21 \, \text{m/s}

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