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Shkiper50 [21]
3 years ago
11

Questions E6 a& b and E7 a&b?

Physics
1 answer:
erica [24]3 years ago
8 0

Explanation:

6a) Work = force × distance

W = Fd

W = (60 N) (10 m)

W = 600 J

6b) Change in energy = work

ΔKE = 600 J

7a) Kinetic energy is half the mass times the square of the velocity.

KE = ½ mv²

KE = ½ (0.4 kg) (25 m/s)²

KE = 125 J

7b) Work = change in energy.  When the ball is stopped, it has zero kinetic energy.

W = ΔKE

W = 0 J − 125 J

W = -125 J

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A car travels across Texas m miles at the rate of t miles per hour. How many hours does the trip take??
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Answer: The trip takes \frac{m}{t}hours

Explanation:

Velocity V is the variation of the position of a body (distance traveled d) with time T:

V=\frac{d}{T}

In this case, the car travels a distance d=m miles at a velocity V=t \frac{miles}{hour} and we need to find the time it takes the trip.

Isolating  T:

T=\frac{d}{V}=\frac{m miles}{t \frac{miles}{hour}}

Finally:

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Read 2 more answers
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

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