A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time interval and gives the cart a final speed. To reach the same speed
using a force that is 1/11 as big, the force must be exerted for a time interval that is ?
since v=at,
and F=ma,
1/11 the force means 11 times as long
To solve this problem, we can use the concept of impulse. The impulse experienced by an object is equal to the product of the force acting on it and the time interval for which the force is applied. Mathematically, impulse (J) is given by the equation:
J = F * Δt
Where J is the impulse, F is the force, and Δt is the time interval.
In this case, we know that the force acts for a short time interval and gives the cart a final speed. Let's call the initial velocity of the cart V_i and the final velocity V_f. Since the cart starts from rest, V_i = 0. We want to find the time interval (Δt) required to reach the same final speed using a force that is 1/11 as big.
We can start by calculating the impulse using the given force. Let's represent the given force as F_1, so the impulse (J_1) is given by:
J_1 = F_1 * Δt
Now, we can use the definition of impulse and the equation for final velocity to relate the impulse with the velocities:
J_1 = m * (V_f - V_i)
Since the mass of the cart (m) is constant and V_i = 0, the equation simplifies to:
J_1 = m * V_f
Next, let's consider the case where the force is 1/11 as big. We can represent this force as F_2. Using the same reasoning, the impulse (J_2) is given by:
J_2 = F_2 * Δt
We want to find the time interval required to reach the same final velocity, so we can set up the equation:
J_2 = m * V_f
Since both situations result in the same impulse (J_1 = J_2), we can equate the two expressions:
F_1 * Δt = F_2 * Δt
Dividing both sides of the equation by Δt, we get:
F_1 = F_2 * (Δt/Δt)
Since F_2 is 1/11 as big as F_1, we can substitute and solve for Δt:
F_1 = (1/11) * F_1 * (Δt/Δt)
1 = (1/11)
Therefore, the time interval (Δt) required to reach the same final speed using a force that is 1/11 as big is equal to 1 second.
We can use the concept of impulse to solve this problem. Impulse can be defined as the product of the force and the time interval over which it is applied. Mathematically, impulse (J) is given by:
J = F * Δt
Where:
J is the impulse
F is the force applied
Δt is the time interval over which the force is applied
In the given scenario, let's assume the initial force (F1) is exerted for a time interval (Δt1) to give the cart a final speed. To reach the same speed using a force that is 1/11 (F2) as big, the force must be exerted for a time interval (Δt2).
According to the impulse-momentum principle, if the initial and final momentum of an object is the same, the impulse experienced by the object will also be the same. Therefore, we can equate the impulses in these two situations:
J1 = J2
F1 * Δt1 = F2 * Δt2
We are given that F2 = (1/11) * F1 (force is 1/11 as big)
Substituting the value of F2 in the equation:
F1 * Δt1 = (1/11) * F1 * Δt2
Now, we can divide both sides of the equation by F1 and solve for Δt2:
Δt2 = Δt1 * 11
Therefore, the time interval (Δt2) over which the smaller force is exerted to reach the same speed as the initial force must be 11 times the time interval (Δt1) over which the initial force is applied.