A car travelling at 25 m/s slams its break leaving skids marks along a distance 125 m as it comes to rest. the coefficient f friction between the tire and the road is
a 5
b 2.5
c 1
2 .4
e .25
A 500N worker pulling a 500 N crate on a rope with a force of 50 N at a constant velocity if the coefficient of friction for the worker is 0.05 and the crate is 0.04 what is the acceleration of the worker?
a .001m/s^2
b .1m/s^2
c 1m/s^2
d 2m/s^2
e 5m/s^2
please show work so i can understand how to solve it. ty in advance.
average speed during stop = 25/2 = 12.5 m/s
time to stop, t = 125/12.5 = 10 s
v = Vi + a t
0 = 25 + a(10)
a = -2.5 m/s^2
F = m a
F = -2.5 m
F = -mu m g
F = -mu m (9.81)
so
mu(9.81)=2.5
mu = 2.5/9.81
I do not understand your second problem.
A car travelling at 25 m/s slams its brakes and leaves skid marks along a distance of 125 m as it comes to rest. To find the coefficient of friction between the tire and the road, we can use the equation:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s), u is the initial velocity (25 m/s), a is the acceleration, and s is the distance (125 m).
Plugging in the values, we get:
0^2 = (25 m/s)^2 + 2a(125 m)
Simplifying:
0 = 625 m^2/s^2 + 250a
250a = -625 m^2/s^2
a = -2.5 m/s^2
Now, since the acceleration is negative, it means the car is decelerating. This can be interpreted as the force of friction acting in the opposite direction to the motion of the car.
The equation for the force of friction is:
f = μN
where f is the force of friction, μ is the coefficient of friction, and N is the normal force. In this case, the normal force is equal to the weight of the car.
Since the weight of the car is not given, we cannot determine the coefficient of friction accurately. Therefore, we cannot choose the correct option. However, if we have an imaginary scale to weigh the car, we can use the weight (500N) to find the coefficient of friction using the formula:
f = μN
f = μ(500N)
We need more information to accurately solve this problem.
Regarding the second question, you mentioned a 500N worker pulling a 500N crate on a rope with a force of 50N at a constant velocity. Since the velocity is constant, the acceleration is zero. Therefore, the answer is not provided in the options you listed. The worker's acceleration is 0 m/s^2.
Remember, you don't always need humor to get an answer. Sometimes, physics can be serious business!
For the first question:
Given:
Initial velocity (u) = 25 m/s
Distance (s) = 125 m
Coefficient of friction (μ) = ?
We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s since the car comes to rest)
u = initial velocity
a = acceleration
s = distance
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (0 - 25^2) / (2 * 125)
a = (-625) / 250
a = -2.5 m/s^2
Since acceleration cannot be negative in this case, the coefficient of friction (μ) should be positive. Therefore, the answer is b) 2.5.
For the second question:
Given:
Force by the worker (Fw) = 50 N
Force of friction (fw) = Fw * μ (where μ is the coefficient of friction for the worker)
Mass of the worker (mw) = Fw / g (where g is the acceleration due to gravity)
Acceleration (a) = ?
Using Newton's second law:
F = ma
Rearranging the equation, we have:
a = F / m
Substituting the known values:
a = (Fw - fw) / mw
a = (50 - (50 * 0.05)) / (500 / 9.8)
a = (50 - 2.5) / (50.98)
a = 47.5 / 50.98
a ≈ 0.93 m/s^2
Therefore, the acceleration of the worker is approximately 0.93 m/s^2.
To solve the first question, we can use the equation F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force.
Given:
Velocity (v) = 25 m/s
Distance (d) = 125 m
To find the coefficient of friction (μ), we first need to calculate the deceleration of the car.
We can use the equation v^2 = u^2 + 2as, where u is the initial velocity, v is the final velocity, a is the acceleration, and s is the distance.
Since the car comes to a stop, the final velocity (v) is 0. The initial velocity (u) is given as 25 m/s, and the distance (s) is given as 125 m.
Substituting the values into the equation, we get:
0 = (25)^2 + 2a(125)
Simplifying the equation:
0 = 625 + 250a
Rearranging the equation:
250a = -625
Solving for acceleration (a):
a = -625 / 250
a = -2.5 m/s^2 (Note: The negative sign indicates deceleration)
Now that we have the acceleration, we can find the normal force (N) acting on the car.
Using the equation F = ma, where F is the force and m is the mass, we can rearrange it to find the normal force:
F = m * a
N = m * a / g, where g is the acceleration due to gravity
Since the mass (m) is not given, we can't directly calculate the normal force. However, we can see that the mass cancels out when calculating the coefficient of friction (μ).
The frictional force (F) can be calculated using the equation F = μN. Since N = m * g, we can rewrite it as F = μmg, where m is the mass and g is the acceleration due to gravity. Again, m cancels out when calculating the coefficient of friction.
Substituting the known values into the equation, we get:
F = μmg
Since the gravitational force (mg) is equal to the weight (W), we can rewrite the equation as:
F = μW
Now, let's find the force (F) required to decelerate the car.
Using the equation F = ma, where F is the force, m is the mass, and a is the acceleration, we can rearrange it to find the force:
F = m * a
Since the mass (m) cancels out when calculating the coefficient of friction, we have:
F = a
Substituting the value for acceleration (a) we found earlier (-2.5 m/s^2), we now have:
F = -2.5
Comparing this with the equation F = μW, we can solve for the coefficient of friction (μ):
F = μW
-2.5 = μW
Since the weight (W) can be calculated using the equation W = mg, where m is the mass and g is the acceleration due to gravity, we can rewrite the equation as:
-2.5 = μmg
The mass (m) cancels out, giving us:
-2.5 = μg
Solving for the coefficient of friction (μ):
μ = -2.5 / g
Now, we can determine the coefficient of friction by using the value for the acceleration due to gravity (g), which is approximately 9.8 m/s^2.
Let's substitute the value into the equation:
μ = -2.5 / 9.8
μ ≈ -0.25
Since the coefficient of friction cannot be negative, we take the absolute value:
μ ≈ 0.25
Therefore, the coefficient of friction between the tire and the road is approximately 0.25. Hence, the correct answer is (e) 0.25.
Now let's solve the second question.
To find the acceleration of the worker, we can use the equation F = μN, where F is the force of friction, μ is the coefficient of friction, and N is the normal force.
Given:
Force of the worker pulling the crate (Fw) = 50 N
Force of friction for the worker (Fwf) = μwNw, where μw is the coefficient of friction for the worker and Nw is the normal force on the worker.
Force of friction for the crate (Fcf) = μcNc, where μc is the coefficient of friction for the crate and Nc is the normal force on the crate.
Total force (Ft) = Fw - Fwf - Fcf, since the forces act in opposite directions due to the constant velocity.
We need to calculate the acceleration (a) of the worker, which can be found using Newton's second law: F = ma.
Equating the forces and substituting for Fw, Fwf, and Fcf, we have:
Ft = Fw - Fwf - Fcf
Ft = ma
Substituting the given values, we get:
50 = ma
Now, let's find the normal forces (Nw and Nc). The normal force is the force exerted perpendicular to a surface, which cancels out the weight of the object.
Since the crate is on a flat surface, the normal force on the worker (Nw) is equal to the weight of the worker (Ww = 500 N). Similarly, we have Nc = Wc = 500 N.
Now, we can find the forces of friction (Fwf and Fcf) for the worker and the crate, respectively. Using the equation F = μN, we can write:
Fwf = μwNw
Fcf = μcNc
Substituting the given values for the coefficients of friction, we have:
Fwf = 0.05 * 500
Fcf = 0.04 * 500
Calculating the forces of friction, we get:
Fwf = 25 N
Fcf = 20 N
Now, let's substitute the values into the equation Ft = ma and solve for acceleration (a):
50 = ma
Rearranging the equation to isolate the acceleration:
a = 50 / m = 50 / (m1 + m2)
Since the masses of the worker (m1) and the crate (m2) are both 500 N, we have:
a = 50 / (500 + 500)
a = 50 / 1000
a = 0.05 m/s^2
Therefore, the acceleration of the worker is 0.05 m/s^2. Hence, the correct answer is (a) 0.001 m/s^2.