A wodden block weighing 200 g is attached to a hanging weighing 45 g with a fine thread . The block is placed on a frictionless surface and after passing thrad over a smooth pulley , hanger is allowed to pull the block . Find the acceleration of the system and tension in the string.
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The answer is
a= 1.8m/s
T=0.36 N
Total mass=200+45=245
F=mg
F=45×10 = 450N
Acceleration(a) = F/total mass =450/245=1.8m/s^2
Tension(T)=ma =200×1.8=360N
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The force pulling the system is .045*9.8 N
That is equal to the acceleration of the system.
a=F/totalmass=.045*9.8/.245 m/s^2
so the tension on the string is the forcepulling on the 200g block...
tension= .200*a=.200*.045*9.8/.245 Newtons
To find the acceleration of the system and the tension in the string, we can use Newton's second law of motion and the concept of force equilibrium.
Let's first analyze the forces acting on the wooden block and the hanging weight.
1. Wooden Block:
- Weight (W_block) = mass (m_block) x acceleration due to gravity (g)
- W_block = 200 g x 9.8 m/s^2 (considering g as 9.8 m/s^2)
- W_block = 1960 g Newtons
2. Hanging Weight:
- Weight (W_hanging) = mass (m_hanging) x acceleration due to gravity (g)
- W_hanging = 45 g x 9.8 m/s^2
- W_hanging = 441 g Newtons
Now, let's consider the forces acting on the system when the hanging weight starts to pull the wooden block.
3. Tension in the String:
- The tension in the string (T) will be the same for both objects since they are connected by a fine thread.
Next, we can set up the equations of motion for both objects using Newton's second law:
1. Wooden Block:
- Net force on the wooden block (F_block) = W_block - T (opposite direction)
- According to Newton's second law, F_block = m_block x acceleration
- W_block - T = m_block x acceleration -------- (Equation 1)
2. Hanging Weight:
- Net force on the hanging weight (F_hanging) = T - W_hanging (opposite direction)
- According to Newton's second law, F_hanging = m_hanging x acceleration
- T - W_hanging = m_hanging x acceleration -------- (Equation 2)
Now, we can solve these two equations simultaneously to find the acceleration (a) and tension in the string (T).
Solving Equation 1 and Equation 2, we get:
1960 g - T = 200 g x a ------- (Equation 3)
T - 441 g = 45 g x a ------- (Equation 4)
Now, substitute the value of T from Equation 3 into Equation 4:
1960 g - (1960 g - 200 g x a) = 441 g + 45 g x a
Simplifying this equation:
200 g x a = 481 g
a = 481 g / 200 g
a = 2.405 m/s^2
Substitute the value of a into Equation 3:
1960 g - T = 200 g x 2.405 m/s^2
T = 1960 g - 481 g
T = 1479 g Newtons
So, the acceleration of the system is 2.405 m/s^2, and the tension in the string is 1479 g Newtons.
the force on the system is the weight (m*g) of the hanging mass
... a = f / m
the tension in the string is the force necessary to give the block the calculated acceleration
... t = m a