Math Problem Statement

A cable with 16.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after being lifted 2.00 m ? Solve this problem using work and energy.

Solution

To solve this problem using work and energy principles, we need to calculate the net work done on the block and then use the work-energy theorem to find the block's speed after it has been lifted 2.00 meters.

Given Information

  • Tension in cable, T=16.0NT = 16.0 \, \text{N}
  • Mass of block, m=1.50kgm = 1.50 \, \text{kg}
  • Distance lifted, d=2.00md = 2.00 \, \text{m}
  • Initial speed, vi=0m/sv_i = 0 \, \text{m/s}
  • Acceleration due to gravity, g=9.81m/s2g = 9.81 \, \text{m/s}^2

Steps to Solve

  1. Calculate the gravitational force acting on the block: Fg=mg=1.50kg×9.81m/s2=14.715NF_g = m \cdot g = 1.50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 14.715 \, \text{N}

  2. Determine the net force acting on the block: Since the block is being lifted straight up, the net force FnetF_{\text{net}} is the difference between the upward tension force and the downward gravitational force: Fnet=TFg=16.0N14.715N=1.285NF_{\text{net}} = T - F_g = 16.0 \, \text{N} - 14.715 \, \text{N} = 1.285 \, \text{N}

  3. Calculate the net work done on the block: The work done by the net force WnetW_{\text{net}} over the distance dd is given by: Wnet=Fnetd=1.285N×2.00m=2.57JW_{\text{net}} = F_{\text{net}} \cdot d = 1.285 \, \text{N} \times 2.00 \, \text{m} = 2.57 \, \text{J}

  4. Apply the work-energy theorem: The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy: Wnet=ΔKE=12mvf212mvi2W_{\text{net}} = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 Since the initial speed viv_i is 0, this simplifies to: Wnet=12mvf2W_{\text{net}} = \frac{1}{2} m v_f^2 Solving for vfv_f, the final speed: vf=2Wnetmv_f = \sqrt{\frac{2 \cdot W_{\text{net}}}{m}}

  5. Substitute the values to find vfv_f: vf=22.57J1.50kgv_f = \sqrt{\frac{2 \cdot 2.57 \, \text{J}}{1.50 \, \text{kg}}} vf=5.141.50=3.42671.85m/sv_f = \sqrt{\frac{5.14}{1.50}} = \sqrt{3.4267} \approx 1.85 \, \text{m/s}

Final Answer

The block's speed after being lifted 2.00 meters is approximately 1.85 m/s.

Would you like more details on any step? Here are some related questions to explore further:

  1. How would the answer change if the block had an initial upward velocity?
  2. What would happen to the block's speed if the tension in the cable was increased?
  3. How would using a variable tension affect the work-energy calculations?
  4. Can you solve this problem using kinematic equations and Newton’s second law instead?
  5. How is gravitational potential energy involved in this work-energy scenario?

Tip: When solving problems using the work-energy theorem, focus on identifying the net force and work done by each force to understand the change in kinetic energy.

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Math Problem Analysis

Mathematical Concepts

Physics
Work-Energy Theorem
Kinetic Energy
Net Force Calculation

Formulas

Gravitational Force: F_g = m * g
Net Work Done: W_net = F_net * d
Work-Energy Theorem: W_net = (1/2) * m * v_f^2 - (1/2) * m * v_i^2
Final Speed Calculation: v_f = sqrt(2 * W_net / m)

Theorems

Work-Energy Theorem

Suitable Grade Level

Grades 10-12