Two blocks of mass 3kg and 2 kg are jointed with a massless and inextensible string. If breaking tension of string is 10N find the maximum horizontal force with two blocks can be pulled on the surface.
50/3N
To find the maximum horizontal force with which the two blocks can be pulled on the surface, we need to analyze the forces acting on the system.
Let's denote the mass of the 3 kg block as m1 and the mass of the 2 kg block as m2.
The forces acting on the 3 kg block are:
- The tension in the string, T, pulling toward the right.
- The force of gravity, mg1, pulling downward (where g is the acceleration due to gravity).
The forces acting on the 2 kg block are:
- The tension in the string, T, pulling toward the left.
- The force of gravity, mg2, pulling downward.
Since the massless and inextensible string transmits the same tension T throughout its length, we can say that T is the same at both ends of the string. Therefore, we can write the following equation:
T = T
Now let's consider the acceleration of the system. Since the system is moving horizontally, the acceleration is the same for both blocks. Let's denote the acceleration as a.
For the 3 kg block, the net force in the horizontal direction is given by:
m1 * a = T
For the 2 kg block, the net force in the horizontal direction is given by:
m2 * a = T
We also know that the breaking tension of the string is 10 N. So the maximum tension that can be safely applied to the string is 10 N.
To find the maximum horizontal force, we need to determine the maximum tension value and then solve for the maximum force. Since the tension is the same on both sides of the string, we can consider the maximum tension value for either block.
Let's consider the 3 kg block:
m1 * a = T
Rearrange the equation:
T = m1 * a
Substitute the given value for T (maximum tension) and m1 (mass of the 3 kg block):
10 N = 3 kg * a
Solve for a (acceleration):
a = 10 N / 3 kg
Substitute the calculated value of a into the equation to find the maximum horizontal force:
F = m2 * a
Substitute the given value for m2 (mass of the 2 kg block):
F = 2 kg * (10 N / 3 kg)
Simplify the equation:
F = 20/3 N
Therefore, the maximum horizontal force with which the two blocks can be pulled on the surface is approximately 6.67 N.
a1 is 10/3 m/s square
a2 is 10/2=5 m/s square
Now
F =(2+3)*5=25 N