Answer to Question #104307 in Classical Mechanics for Jeffrey Chen

Question

1. I drag a mass m = 21 kg in a straight line, along a horizontal surface, a distance D = 23 m . I drag it at constant speed v = 0.90 m.s^-1 in a straight line using a horizontal force. The coefficients of friction are μs = 1.2 and μk = 1.1. How much work do I do?

2. A rubber band has mass m = 0.30 g and a spring constant k =15 N.m^-1 . I stretch it by 5.0 cm (which in this case doubles its length). Assume the rubber band behaves as a Hooke's law spring. Assume that, when you launch the rubber band, all of the stored potential energy is converted into kinetic energy. How fast is it at the launch?

3. A car has a drag coefficient Cd = 0.30, a frontal area of A = 1.9 m^2 and a mass 1.2 tonnes. The density of air is 1.2 kg.m^-3. Hint: Retain accurate values until the final calculation, but remember significant figures.

i) What is the drag force when it is traveling at v = 110 kph in a straight line?
ii) What power (in kilowatts) is required to overcome the drag force at this speed?

Answer 1

As per the given question,

Given mass = 21 kg

Distance on the horizontal surface (d) = 23m

Speed = 0.90 m/sec

Coefficient of friction $\mu_s =1.2$

$\mu_k =1.1$

Friction force = $f_s=\mu mg=1.1\times21\times9.8=226.38N$

work due to friction =$f_s . d=226.38\times23=5206.74J$

Kinetic energy of the object = $\dfrac{mv^2}{2}=\dfrac{21\times 0.9^2}{2}=8.505J$

Hence net work = $8.505+226.38 =234.885J$

2.

As per the given question,

mass (m)=0.3 g=$3\times 10^{-4} kg$

Spring constant (k) =$15 N/m$

Stretched length$(\Delta x)=5.0 \times 10^{-2}m$

Let it will move with the velocity v,

So applying the conservation of the energy,

$\dfrac{mv^2}{2}=\dfrac{Kx^2}{2}$

$\Rightarrow v=\sqrt{\dfrac{K}{m}}x$

$\Rightarrow v=\sqrt{\dfrac{15}{0.3\times 10^{-4}}}\times 5\times 10^{-2} m/sec$

$\Rightarrow v=7.07\times10^{2}\times 5\times10^{-2}=35.35J$