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stellarik [79]
3 years ago
11

A block of mass 9.5kg rests on a slope an angle of 23.0∘ relative to the horizontal. What is the size of the contact force norma

l to the slope, in Newtons?
Use a gravitational acceleration value of g = 9.8m/s^2.
Physics
1 answer:
Rudiy273 years ago
6 0

The Normal Force = M x G x Cos(theta)

= 9.5 Kg x 9.8 m/s^2 x cos 23

= 9.5 Kg x 9.8 m/s^2 x 0.9205

Converting Kg to Newton,

1 Kg  = 9.81 N

= 9.5 Kg x 9.81 N x 9.8 m/s^2 x 0.9205

= 840.702 N

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The person has an upward acceleration of 0.70 m/s^2 and is lifted from rest through a distance of 13 m. A rescue helicopter lift
Alex

Answer:

a)

840 N

b)

10920 J

c)

- 10192 J

d)

4.3 m/s

Explanation:

a)

T = tension force in the cable in upward direction = ?

a = acceleration of the person in upward direction = 0.70 m/s²

m = mass of the person being lifted = 80 kg

Force equation for the motion of person in upward direction is given as

T - mg = ma

T = m (g + a)

T = (80) (9.8 + 0.70)

T = 840 N

b)

d = distance traveled in upward direction = 13 m

W_{t} = Work done by tension force

Work done by tension force is given as

W_{t} = T d

W_{t} = (840) (13)

W_{t} = 10920 J

c)

d = distance traveled in upward direction = 13 m

W_{g} = Work done by person's weight

Work done by person's weight is given as

W_{g} = - mg d

W_{g} = - (80 x 9.8) (13)

W_{g} = - 10192 J

d)

F_{net} = Net force on the person = ma = 80 x 0.70 = 56 N

v₀ = initial speed of the person = 0 m/s

v = final speed

Using work-energy theorem

F_{net} d = (0.5) m (v² - v₀²)

(56) (13) = (0.5) (80) (v² - 0²)

v = 4.3 m/s

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4 years ago
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I think the answer is D.
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I’m pretty sure the answer is A
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