Calculate the kinetic energy of a 20-kg sled moving 28.0 m/s. Show your work in the space to the right.

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

nirvana33 [79]

Answer:

a) $F = 882.63\,N$ , b) $\dot W= 4413.15\,W$ , c) $\eta = 15.216\,\%$ .

Explanation:

a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:

$\Sigma F_{x} = F - \mu_{r}\cdot N = 0$

$\Sigma F_{y} = N - m\cdot g = 0$

After some algebraic handling, the following expression for the propulsion force is constructed:

$F = \mu_{r}\cdot m \cdot g$

$F = (0.06)\cdot (1500\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 882.63\,N$

b) The power require to move the car at a speed of 5 meters per second is:

$\dot W = F\cdot v$

$\dot W = (882.63\,N)\cdot (5\,\frac{m}{s} )$

$\dot W= 4413.15\,W$

c) The efficiency of the car is:

$\eta = \frac{(882.63\,N)\cdot (15\,mi)\cdot (\frac{1609\,m}{1\,mi} )}{(1.4\times 10^{8}\,J)} \times 100\,\%$

$\eta = 15.216\,\%$

3 0

3 years ago

At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people o

velikii [3]

Let s = rate of rotation
Let r = radius of earth = 6,400km 
Then solving (s^2) r = g will give the desired rate, from which length of day is inferred. 
People would not be thrown off. They would simply move eastward in a straight line while the curved surface of earth fell away from beneath them.

5 0

3 years ago

Read 2 more answers

2)

earnstyle [38]

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

$x=0.150m-0.115m=0.035m$

And spring constant is $k=51.0N/m$

$PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312$

Hence, the potential energy is 0.0312J

8 0

3 years ago

Which of the following is a terrestrial planet on which one would observe the Sun rising in the west and setting in the east?

Romashka-Z-Leto [24]

The correct answer is Venus

4 0

3 years ago

Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I pr

Sindrei [870]

Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the Principle of Conservation of Energy, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".

When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.

So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.

The representation is as followed:

$K_{i}+P_{i}=K_{f}+P_{f}$

$K_{i}+0=0+P_{f}$

$\frac{1}{2}mv^{2} = mgh$

As we noticed, mass of the object can be cancelled from the equation, making height be:

$h=\frac{v^{2}}{2g}$

So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.

6 0

2 years ago