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babymother [125]
3 years ago
12

Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp

lain and show your work. First solve for work. Then solve for power.
Physics
1 answer:
shutvik [7]3 years ago
3 0

Answer: 500 Watts

Explanation:

Power P is the speed with which work W is done. Its unit is Watts (W), being 1 W=\frac{1 Joule}{1 s}.

Power is mathematically expressed as:

P=\frac{W}{t} (1)

Where t is the time during which work W  is performed.

On the other hand, the Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter (1J=(1N)(1m)=Nm  ).

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (2)

In this case, we have the following data:

F=500 N

d=10 m

t=10 s

So, let's calculate the work done by Peter and then find how much power is involved:

From (2):

W=(500 N)(10 m) (3)

W=5000 J (4)

Substituting (4) in (1):

P=\frac{5000 J}{10 s} (5)

Finally:

P=500 W

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Answer:

Explanation:

Given that,

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M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

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v(t) = -45.5 Sin(3.922×0.152)

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v(t) = -25.5 cm/s

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a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

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