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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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1) The equivalent resistance of two resistors in parallel is given by:
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so in our problem we have
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natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t

where,

v₀ = initial speed = 110 ft/s

Therefore,

110 = 32t\\\\t = \frac{110}{32}\\\\

<u>t = 3.48 s</u>

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