A bicyclist was moving at a rate of 5 m/s, and then sped up to 9 m/s. If the cyclist has a mass of 130 kg, how much work was needed to increase his velocity?

Question

Layla Ahmed · Accepted Answer

#3640J#

We know that KE#=1/2mv^2# ......(1)

Therefore, Change in KE, #Delta#KE#= 1/2 mv_f^2-1/2 mv_i^2# #=>Delta#KE#= 1/2 m(v_f^2-v_i^2)# ......(2) where #v_iand v_f# are initial and final velocities respectively of the cyclist.

Plug in all this information into the equation (2) #Delta#KE#= 1/2 xx130(9^2-5^2)# #=>Delta#KE#= 3640J#

Assuming that there are no losses, the change in kinetic energy is equal to the work done.

Adam Arnold · Answer

undefined The work needed to increase the velocity of the cyclist can be calculated using the work-energy principle. The change in kinetic energy is equal to the work done.

Initial kinetic energy $KE_1 = \frac{1}{2} mv_1^2$  
Final kinetic energy $KE_2 = \frac{1}{2} mv_2^2$

Change in kinetic energy $ΔKE = KE_2 - KE_1$  
Work done = $ΔKE$

Given:  
Initial velocity $v_1 = 5 \, \text{m/s}$  
Final velocity $v_2 = 9 \, \text{m/s}$  
Mass of the cyclist $m = 130 \, \text{kg}$

Calculate $ΔKE = KE_2 - KE_1$  
$KE_1 = \frac{1}{2} \times 130 \times (5)^2$  
$KE_2 = \frac{1}{2} \times 130 \times (9)^2$

$ΔKE = \frac{1}{2} \times 130 \times (9)^2 - \frac{1}{2} \times 130 \times (5)^2$

$ΔKE = \frac{1}{2} \times 130 \times (81 - 25)$

$ΔKE = \frac{1}{2} \times 130 \times 56$

$ΔKE = 3640 \, \text{J}$

Therefore, the work needed to increase the cyclist's velocity is $3640 \, \text{J}$.