Question

Use the approximation that g ≈ 10 m/s2 to fill in the blanks in the following diagram. Ftension (force of tension) = ___________ N asystem (acceleration of the system) = ___________ m/s/s ΣFsystem (net force of the system) = ___________ N

Answer 1

We are asked to determine the tension in the system. To do that we will add the vertical forces on the 50g mass using the following free-body diagram:

Where:

$\begin{gathered} T=\text{ tension} \\ m=50g\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}$

Adding the vertical forces we get (taking downward forces to be positive and upward negative):

$mg-T=ma$

Where "a" is the acceleration of the system.

If we divide by mass "m" we get:

$\frac{mg-T}{m}=a$

Now, we add the horizontal forces for the 250g mass using the following free-body diagram:

Where:

$\begin{gathered} T=\text{ tension} \\ M=250g\text{ mass} \\ F_f=\text{ friction force} \\ N=\text{ normal force} \end{gathered}$

Now, we add the horizontal forces we get:

$T-F_f=Ma$

The friction force is given by:

$F_f=\mu N$

Since the object is not accelerating in the vertical direction this means that the normal force is equivalent to the weight of the 250g mass, therefore, we have:

$F_f=\mu Mg$

Substituting in the sum of forces we get:

$T-\mu Mg=Ma$

Now, we divide both sides by the mass "M":

$\frac{T-\mu Mg}{M}=a$

Since the acceleration is the same for both masses we can set them equal together and we get:

$\frac{mg-T}{m}=\frac{T-\mu Mg}{M}$

Now, we solve for the tension "T". First, we distribute both denominators:

$g-\frac{T}{m}=\frac{T}{M}-\mu g$

Now, we add T/m:

$g=\frac{T}{M}+\frac{T}{m}-\mu g$

Now, we add the coefficient of friction and the gravity to both sides:

$g+\mu g=\frac{T}{M}+\frac{T}{m}$

Now, we take "T" as a common factor:

$g+\mu g=T(\frac{1}{M}+\frac{1}{m})$

Now, we divide both sides by the factor:

$\frac{g+\mu g}{\frac{1}{M}+\frac{1}{m}}=T$

Now, we substitute the values:

$\frac{9.8\frac{m}{s^2}+0.1(9.8\frac{m}{s^2})}{\frac{1}{0.25kg}+\frac{1}{0.05kg}}=T$

Solving the operations:

$0.45N=T$

Therefore, the force of tension is 0.45 N.

Now, we determine the acceleration by substituting the value of the force of tension in any of the two formulas for acceleration:

$a=\frac{mg-T}{m}$

Substituting the values:

$a=\frac{(0.05kg)(9.8\frac{m}{s^2})-0.45N}{0.05kg}$

Solving the operations:

$a=0.8\frac{m}{s^2}$

Therefore, the acceleration is 0.8 m/s/s.

The net force on the system is given by Newton's second law:

$\Sigma F_{net}=(M+m)a$

Where "M + m" is the total mass of the system and "a" is the acceleration of the system. Substituting the values we get:

$\Sigma F_{net}=(0.25kg+0.05kg)(0.8\frac{m}{s^2})$

Solving the operations:

$\Sigma F_{net}=0.24N$

Therefore, the net force is 0.24 N.