A mom is her daughter in a stroller . The mass of the stroller is 65kg The mom is exerting a force of 180N at an angle of 30 degrees . If the coefficient of friction between the wheels and the ground is 0.285 , find the acceleration of the stroller
The Correct Answer and Explanation is:
Let’s solve this problem step by step.
Step 1: Identify the forces acting on the stroller
- Force exerted by the mom: This is 180 N at an angle of 30 degrees. We need to break this force into two components:
- Horizontal component (FhorizontalF_{\text{horizontal}}Fhorizontal) = 180×cos(30∘)180 \times \cos(30^\circ)180×cos(30∘)
- Vertical component (FverticalF_{\text{vertical}}Fvertical) = 180×sin(30∘)180 \times \sin(30^\circ)180×sin(30∘)
- Frictional force: The frictional force ffrictionf_{\text{friction}}ffriction is given by: ffriction=μ×Nf_{\text{friction}} = \mu \times Nffriction=μ×N where μ\muμ is the coefficient of friction (0.285), and NNN is the normal force. The normal force is the force exerted by the ground on the stroller, balancing the weight and the vertical component of the mom’s force. The normal force NNN is given by: N=mg−FverticalN = mg – F_{\text{vertical}}N=mg−Fvertical where mmm is the mass of the stroller (65 kg) and ggg is the acceleration due to gravity (9.8 m/s²).
Step 2: Calculate the components
- Horizontal force: Fhorizontal=180×cos(30∘)≈180×0.866≈155.88 NF_{\text{horizontal}} = 180 \times \cos(30^\circ) \approx 180 \times 0.866 \approx 155.88 \, \text{N}Fhorizontal=180×cos(30∘)≈180×0.866≈155.88N
- Vertical force: Fvertical=180×sin(30∘)=180×0.5=90 NF_{\text{vertical}} = 180 \times \sin(30^\circ) = 180 \times 0.5 = 90 \, \text{N}Fvertical=180×sin(30∘)=180×0.5=90N
- Normal force:
The weight of the stroller is mg=65×9.8=637 Nmg = 65 \times 9.8 = 637 \, \text{N}mg=65×9.8=637N. So, the normal force is: N=637−90=547 NN = 637 – 90 = 547 \, \text{N}N=637−90=547N - Frictional force: ffriction=0.285×547≈155.85 Nf_{\text{friction}} = 0.285 \times 547 \approx 155.85 \, \text{N}ffriction=0.285×547≈155.85N
Step 3: Calculate the net horizontal force
The net horizontal force is the difference between the horizontal component of the force exerted by the mom and the frictional force: Fnet=Fhorizontal−ffriction=155.88−155.85≈0.03 NF_{\text{net}} = F_{\text{horizontal}} – f_{\text{friction}} = 155.88 – 155.85 \approx 0.03 \, \text{N}Fnet=Fhorizontal−ffriction=155.88−155.85≈0.03N
Step 4: Calculate the acceleration
Now, we can use Newton’s second law to find the acceleration of the stroller: Fnet=maF_{\text{net}} = maFnet=ma
where m=65 kgm = 65 \, \text{kg}m=65kg is the mass of the stroller. Solving for aaa: a=Fnetm=0.0365≈0.00046 m/s2a = \frac{F_{\text{net}}}{m} = \frac{0.03}{65} \approx 0.00046 \, \text{m/s}^2a=mFnet=650.03≈0.00046m/s2
Conclusion:
The acceleration of the stroller is approximately 0.00046 m/s20.00046 \, \text{m/s}^20.00046m/s2. The very small acceleration is due to the fact that the frictional force almost completely cancels out the force exerted by the mom.
