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VLD [36.1K]
2 years ago
15

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

on the circle, the speed of the mass is 12 m/s. What is the magnitude of the force of the string on the mass at this position?
Physics
2 answers:
il63 [147K]2 years ago
7 0

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

brilliants [131]2 years ago
3 0

Answer: 41N

Explanation :

T= mv^2/R + mgcos θ

At the highest point on the circle θ=0

Cos 0 = 1

T= mv^2/R + mg

m = 0.5kg

Velocity at the highest point (amplitude)= 12m/s

T = 0.5× 12^2/2 + 0.5×10

0.5×144/2 +5

T = 0.5×72 + 5

T = 36+5

T = 41N

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2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting po
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a) t = 11.2 s

b) v = 70.5 mph

Explanation:

a)

  • Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:

       t = \frac{v_{f} - v_{o}}{a}  (1)

  • where vf = 50 mph, and v₀ = 10 mph.
  • However, we still lack the value of a.
  • Assuming that the acceleration is constant, we can use the following kinematic equation:

       v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x  (2)

  • Since we know that Δx = 500 ft, we could solve (2) for a.
  • In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:

       v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s  (3)

       v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s  (4)

  • We can do the same process with Δx, from ft to m, as follows:

       \Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m  (5)

  • Replacing (3), (4), and (5) in (2) and solving for a, we get:

       a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} =  \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m}  = 1.6 m/s2  (6)

  • Replacing (6) in (1) we finally get the value of the time t:

        t = \frac{v_{f} - v_{o}}{a} =  \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2}  = 11.2 s  (7)

b)

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       v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)

  • If we convert vf again to mph, we have:

       v_{f} = 31.5m/s*\frac{1mi}{1609m} *\frac{3600s}{1h} = 70.5 mph  (9)

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