If a #12 kg# object is constantly accelerated from #0m/s# to #60 m/s# over 8 s, how much power musy be applied at #t=1 #?

Question

Leo Abeyta · Accepted Answer

The power is #=675W#

We apply the equation of motion

#v=u+at#

The initial velocity is #u=0 ms^-1#

The final velocity is #u=60 ms^-1#

The time is #t=8s#

The acceleration is

#a=(v-u)/t=(60-0)/8=7.5ms^-1#

The mass is #=12kg#

According to Newton's Second Law, the force is

#F=ma=12*7.5=90N#

The velocity after #t=1s# is

#v=0+7.5*1=7.5ms^-1#

The power is

#P=Fv=90*7.5=675W#

Carson Alexander · Answer

undefined To determine the power at t=1, we need to find the acceleration of the object at that moment. We can use the formula for acceleration: $a = \frac{\Delta v}{\Delta t}$. Plugging in the given values: $a = \frac{60 \, \text{m/s} - 0 \, \text{m/s}}{8 \, \text{s}} = 7.5 \, \text{m/s}^2$. Then, we can use the formula for power: $P = F \times v$, where $F$ is the force and $v$ is the velocity. Since force is equal to mass times acceleration ($F = m \times a$), we have $F = 12 \, \text{kg} \times 7.5 \, \text{m/s}^2 = 90 \, \text{N}$. Finally, substituting $F$ and $v$ into the power formula, we get $P = 90 \, \text{N} \times 0 \, \text{m/s} = 0 \, \text{W}$. Therefore, no power is required at $t=1$.