Question

A 15 kgblock is accelerated at$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$along a horizontal frictionless surface, with the speed increasing from 10 m/sto 30 m/s. What are (a) the change in the block’s mechanical energy and (b) the average rate at which energy is transferred to the block? What is the instantaneous rate of that transfer when the block’s speed is (c) 10 m/sand (d) 30 m/s?

Short answer

a) The block’s mechanical energy is$∆E_{th}=6000J$

b) The average rate at which energy is transferred to the block is$P_{avg}=600W$

c) The instantaneous rate of transfer at$v_i=10m/s$ is$P_{10}=300W$

d) The instantaneous rate of transfer at$v_f=30m/s$ is$P_{30}=900W$

Step-by-step solution

Given

The mass of the block is m = 15 kg

The acceleration of the block is$a=2.0m/s^2$

The initial speed of the block is$v_i=10m/s$

The final speed of the block is$v_f=30m/s$

To understand the concept

As the system is isolated and moving horizontally, the change in mechanical energy is equal to the change in kinetic energy. Also, by using the formula for average power and instantaneous power the required values can be calculated.

(a) Calculate the change in the block’s mechanical energy

Mechanical energy can be written as

$$\begin{gathered} ∆E_{mech}=∆K \\ {K}_f=K_i \\ =\frac{1}{2i}m\left(v_f^2-v_i^2\right) \end{gathered}$$

Using the given values intheabove equation, we have,

$$\begin{gathered} ∆E_{mech}=\frac{1}{2}\times 15kg\left({\left(30m/s\right)}^2-{\left(10m/s\right)}^2\right) \\ =\frac{1}{2}\times 15kg\left(800m^2/s^2\right) \\ =6000kg.m^2/s^2\left(\frac{1J}{1kg.m^{2/s^2}}\right) \\ =6000J \end{gathered}$$

(b) Calculate the average rate at which energy is transferred to the block

Average power is given by

$P_{avg}=\frac{∆E}{∆t}$

Using the kinetic equation,

$$\begin{gathered} V_f=V_i+at \\ t=\frac{V_f-V_i}{a} \\ t=\frac{30m/s-10m/s}{2m/s^2} \\ t=10s \end{gathered}$$

Using the value of$∆E_{mech}$in above equation for average power,

$$\begin{gathered} P_{avg}=\frac{6000J}{10s} \\ =600J/s\left(\frac{1W}{1J/s}\right) \\ =600W \end{gathered}$$

(c) Calculate the instantaneous rate of that transfer when the block’s speed is 10 m/s

The instantaneous rate of the transfer is expressed as,

$\mathrm{P}=\mathrm{F}\mathrm{vcos\theta }$

Since the value of$\cos \theta =1$

For the initial speed of 10 m/s, the instantaneous transfer can be expressed as,

$$\begin{gathered} P=F\times v \\ {P}_{10}=F\times v_i \\ F=ma \\ =15kg\times 2m/s^2 \\ =30kg.m/s^2\left(\frac{1N}{1kg.m/s^2}\right) \\ =30N \end{gathered}$$

Using the value of F and initial velocity in equation (2), we get,

$$\begin{gathered} P_{10}=30N\times 10m/s \\ =300N.m/s\left(\frac{1W}{1N.m/s}\right) \\ =300W \end{gathered}$$

(d) Calculate the instantaneous rate of that transfer when the block’s speed is 30 m/s

For the final speed of 30 m/s, the instantaneous transfer is,

${\mathrm{P}}_{30}=\mathrm{F}\times {\mathrm{v}}_{\mathrm{f}}$

Using the value of F and final velocity, we get,

role="math" localid="1661405296794" $$\begin{gathered} P_{30}=30N\times 30m/s \\ =900N.m./s\left(\frac{1w}{1N.m/s}\right) \\ =900W \end{gathered}$$