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Aug 14, 2024
Solution by Steps
step 1
To find the work done due to friction, we can use the work-energy principle, which states that the work done is equal to the change in kinetic energy. The formula for work done W W is given by: W=ΔKE=KEB−KEA W = \Delta KE = KE_B - KE_A where KE=12mv2 KE = \frac{1}{2}mv^2
step 2
Calculate the initial kinetic energy at point A: KEA=12×1000 kg×(10 ms−1)2=50000 J KE_A = \frac{1}{2} \times 1000 \, \text{kg} \times (10 \, \text{ms}^{-1})^2 = 50000 \, \text{J}
step 3
Calculate the final kinetic energy at point B: KEB=12×1000 kg×(14 ms−1)2=98000 J KE_B = \frac{1}{2} \times 1000 \, \text{kg} \times (14 \, \text{ms}^{-1})^2 = 98000 \, \text{J}
step 4
Now, calculate the work done: W=KEB−KEA=98000 J−50000 J=48000 J W = KE_B - KE_A = 98000 \, \text{J} - 50000 \, \text{J} = 48000 \, \text{J} . Since the work done against friction is negative, we have Wfriction=−48000 J W_{\text{friction}} = -48000 \, \text{J}
step 5
To find the frictional force Ff F_f , we can use the formula W=Ff⋅d W = F_f \cdot d , where d d is the distance traveled (200 m). Rearranging gives Ff=Wd=−48000 J200 m=−240 N F_f = \frac{W}{d} = \frac{-48000 \, \text{J}}{200 \, \text{m}} = -240 \, \text{N}
Answer
Work done due to friction is -48000 J; Frictional force is -240 N
Key Concept
The work-energy principle relates the work done on an object to its change in kinetic energy.
Explanation
The work done against friction is negative, indicating energy is lost due to friction, and the frictional force can be calculated using the work done over the distance.
Solution by Steps
step 2
Calculating Wdown W_{\text{down}} : Wdown=452×9.81×0.2419≈1070 N W_{\text{down}} = 452 \times 9.81 \times 0.2419 \approx 1070 \, \text{N}
step 3
To calculate the tension in the cable, we use the equation: T=Wdown+Ffriction+ma T = W_{\text{down}} + F_{\text{friction}} + ma , where Ffriction=525 N F_{\text{friction}} = 525 \, \text{N} and a=0.13 ms−2 a = 0.13 \, \text{ms}^{-2} . Thus, T=1070+525+(452×0.13) T = 1070 + 525 + (452 \times 0.13)
step 4
Calculating T T : T=1070+525+58.76≈1653.76 N T = 1070 + 525 + 58.76 \approx 1653.76 \, \text{N} . Therefore, T≈1650 N T \approx 1650 \, \text{N}
step 5
To find the time taken to move from S S to P P , we use the formula: s=ut+12at s = ut + \frac{1}{2}at , where u=0 u = 0 (initial velocity), s=10.0 m s = 10.0 \, \text{m} , and a=0.13 ms−2 a = 0.13 \, \text{ms}^{-2} . Thus, 10=0+12×0.13×t2 10 = 0 + \frac{1}{2} \times 0.13 \times t^2
step 6
Rearranging gives t2=10×20.13≈153.846 t^2 = \frac{10 \times 2}{0.13} \approx 153.846 , so t≈153.846≈12.4 s t \approx \sqrt{153.846} \approx 12.4 \, \text{s}
step 7
To find the magnitude of the velocity at P P , we use v=u+at v = u + at . Thus, v=0+0.13×12.4≈1.612 ms−1 v = 0 + 0.13 \times 12.4 \approx 1.612 \, \text{ms}^{-1} , which rounds to 1.6 ms−1 1.6 \, \text{ms}^{-1}
[1] Answer
A
Key Concept
Newton's second law and kinematics
Explanation
The solution involves applying Newton's second law to find tension and using kinematic equations to determine time and velocity.
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