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2 years ago

. A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the

mass acquires a velocity of 4 ms-1 in 8s [g= 10ms-2]

Physics

1 answer:

telo118 [61]2 years ago

5 0

$\text{Given that,}\\\\\text{Mass, m =10 kg}\\\\\text{Time, t = 8 sec}\\\\\text{Velocity, v = 4~m/s}\\\\\text{When a body is moving upwards,}\\\\\text{Tension,}~ T=mg +ma\\\\~~~~~~~~~~~~~~~=mg+m\left(\dfrac{v-u}t \right)\\\\~~~~~~~~~~~~~~~=10(10)+10\left(\dfrac{4-0}8\right)\\\\~~~~~~~~~~~~~~~=100+10\left(\dfrac 12\right)\\\\~~~~~~~~~~~~~~~=100+5\\\\~~~~~~~~~~~~~~~=105~N$

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Answer:

0.976 c

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3 years ago

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3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

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3 years ago