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Review Conceptual Example 16 as background for this problem. The water skier there has a mass of 73 kg. Find the magnitude of the net force acting on the skier when (a) she is accelerated from rest to a speed of 11 m/s in 8.0 s and (b) she lets go of the tow rope and glides to a halt in 21 s.

User Fatouma
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2 Answers

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Final answer:

To calculate the net force on the water skier during acceleration, use Newton's second law of motion, and for deceleration, use the same law with the consideration that the final velocity is zero. The net force during acceleration is 100.375 N, and the force during deceleration is 38.2374 N in magnitude.

Step-by-step explanation:

The student's question involves calculating the magnitude of the net force acting on a water skier during two different phases of motion, applying Newton's second law of motion.

Part (a) Acceleration Phase

To find the magnitude of the net force when the skier accelerates from rest to a speed of 11 m/s in 8.0 s, we can use the formula F = ma, where m is the mass and a is the acceleration. First, we calculate the acceleration as a = ∆v/∆t, where ∆v is the change in velocity and ∆t is the time taken. The change in velocity (∆v) is 11 m/s and the time (∆t) is 8.0 s, giving us an acceleration of 1.375 m/s². With the skier's mass of 73 kg, the net force is F = 73 kg × 1.375 m/s² = 100.375 N.

Part (b) Deceleration Phase

When the skier lets go of the tow rope and glides to a halt in 21 s, we can assume a uniform deceleration. We calculate the deceleration using the same formula for acceleration with the final velocity being 0 m/s. This gives us a deceleration of -0.5238 m/s² (negative sign indicates deceleration). The magnitude of the net force during deceleration is then F = 73 kg × -0.5238 m/s² = -38.2374 N, and since force is a vector quantity, we are only interested in the magnitude, which is 38.2374 N.

User Bero
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Answer:

a) F = 100.375 N

b) F = 38.252 N

Step-by-step explanation:

a) she is accelerated from rest to a speed of 11 m/s in 8.0 s

We get the acceleration we use this equation,

v = u + at

where v is the final velocity - 11 m/s

u = initial velocity = 0 m/s (from rest)

t is time taken = 8 s

Since v = u + at

11 = 0 + a×8

8a = 11

a = 11/8 = 1.375

a = 1.375 m/s²

The force equation is given by:

F = ma

F is the force, m is mass = 73 kg and a is acceleration

F = 73 × 1.375 = 100.375

F = 100.375 N

(b) she lets go of the tow rope and glides to a halt in 21 s.

We get the acceleration we use this equation,

v = u + at

where v is the final velocity = 0 m/s (to a halt)

u = initial velocity = 11 m/s

t is time taken = 21 s

Since v = u + at

0 = 11 + a×21

21a = -11

a = -11/21 = -0.524

a = 0.524 m/s²

The force equation is given by:

F = ma

F is the force, m is mass = 73 kg and a is acceleration

F = 73 × 0.524 = 38.252

F = 38.252 N

User Lukas Lechner
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