a toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a breaking strength of 40 N. Calculate the maximum speed the cart can attain without breaking the string solution
3.7
Well, well, well, looks like we have ourselves a daring circus performer here – a cart on a string! Let's find out just how fast this little daredevil can go without breaking the string.
To figure out the maximum speed, we need to consider the force acting on the cart. In this case, the force causing tension in the string is provided by the centripetal force:
F = m * v^2 / r,
where F is the tension force, m is the mass of the cart, v is its speed, and r is the radius (string length).
Since we want to find the maximum speed without breaking the string, we need to set the tension force equal to the breaking strength of the string:
F = 40 N.
Now, we can rearrange the centripetal force equation to solve for v:
v^2 = F * r / m,
v = sqrt(F * r / m).
Let's plug in the values we have:
v = sqrt((40 N) * (0.70 m) / (2.0 kg)).
Calculating it out, we get:
v ≈ 4.98 m/s.
So, there you have it! The maximum speed the cart can attain without pulling off a "string break" dance is approximately 4.98 meters per second. Just be careful not to go faster or you might unleash the wrath of the string!
To find the maximum speed the cart can attain without breaking the string, we need to consider the tension in the string when the cart is at its maximum speed.
The centripetal force required to keep an object moving in a circular path is given by the equation:
Fc = (mv^2) / r
Where Fc is the centripetal force, m is the mass of the cart, v is the velocity, and r is the radius (length of the string). In this case, m = 2.0 kg and r = 0.70 m.
The tension in the string must be equal to or less than the breaking strength of 40 N. The tension in the string can be calculated using the equation:
Tension = Fc + mg
Where g is the acceleration due to gravity, approximately 9.8 m/s^2.
Setting the tension equal to the breaking strength, we can solve for v:
40 N = [(2.0 kg)(v^2) / 0.70 m] + [(2.0 kg)(9.8 m/s^2)]
Rearranging the equation, we get:
(v^2) = (40 N - (2.0 kg)(9.8 m/s^2)) * (0.70 m) / (2.0 kg)
Simplifying the equation, we find:
(v^2) = 25.9 m^2/s^2
Finally, taking the square root of both sides to solve for v, we find:
v = √(25.9 m^2/s^2)
v ≈ 5.09 m/s
Therefore, the maximum speed the cart can attain without breaking the string is approximately 5.09 m/s.
To calculate the maximum speed the cart can attain without breaking the string, we need to consider the centripetal force acting on the cart.
Centripetal force is the force that keeps an object moving in a circular path. It is given by the equation:
F = m * v^2 / r
Where:
F is the centripetal force,
m is the mass of the object (cart),
v is the velocity (speed) of the object, and
r is the radius of the circular path (string length).
In this case, the maximum speed is reached when the centripetal force equals the breaking strength of the string.
Using the given values:
m = 2.0 kg (mass of the cart),
r = 0.70 m (length of the string), and
F = 40 N (breaking strength of the string),
We can rearrange the formula to solve for v:
v = sqrt(F * r / m)
Plugging in the values, we get:
v = sqrt(40 N * 0.70 m / 2.0 kg)
v = sqrt(28 N*m / 2.0 kg)
v = sqrt(14 m^2/s^2)
v ≈ 3.74 m/s
Therefore, the maximum speed the cart can attain without breaking the string is approximately 3.74 m/s.
Centripetal force = M*V^2/R = 40 N
Solve for V
R is the radius im meters and M is the mass in kg.