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valkas [14]
2 years ago
9

A cable with 19.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after b

eing lifted 2.00 m ? Solve this problem using work and energy
Physics
1 answer:
frez [133]2 years ago
7 0

The final speed of the block, after being lifted 2.00 m is 3.39 m/s

<h3>What is speed?</h3>

Speed can be defined as the rate of change in the distance of a body.

To calculate the speed of the block after being lifted 2.00 m,  first, we need to calculate the acceleration of the block using the formula below

Formula:

  • T-mg = ma......... Equation 1

Where:

  • T = Tension in the cable
  • m = mass of the cable
  • a = acceleration
  • g = acceleration due to gravity

Restructuring the formula above,

  • a = (T-mg)/m............... Equation 2

From the question,

Given:

  • T = 19 N
  • m = 1.5 kg
  • g = 9.8 m/s²

Substitute these values into equation 2

  • a = [(19)-(1.5×9.8)]/1.5
  • a = 4.3/1.5
  • a = 2.87 m/s²

Finally, to calculate the speed of the block, we use the formula below.

  • v² = u²+2as.......... Equation 3

Where:

  • v = Final speed
  • u = initial speed
  • a = acceleration
  • s = distance

From the question,

Given:

  • u = 0 m/s
  • a = 2.87 m/s²
  • s = 2.00 m

Substitute these values into equation 3

  • v² = 0²+(2×2×2.87)
  • v² = 11.48
  • v = √11.48
  • v = 3.39 m/s

Hence, The final speed of the block, after being lifted 2.00 m is 3.39 m/s.

Learn more about speed here: brainly.com/question/6504879

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Read 2 more answers
A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f
Setler [38]

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

W_f = F_fs = 15*0.5 = 7.5 J

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

E_p = mgh = 0.1*10*1.5 = 1.5 J

Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

6 0
3 years ago
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