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:

`(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":

`(T-\mu Mg)/(M)=a`

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

`(mg-T)/(m)=(T-\mu Mg)/(M)`

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

`g-(T)/(m)=(T)/(M)-\mu g`

Now, we add T/m:

`g=(T)/(M)+(T)/(m)-\mu g`

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

`g+\mu g=(T)/(M)+(T)/(m)`

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

`g+\mu g=T((1)/(M)+(1)/(m))`

Now, we divide both sides by the factor:

`(g+\mu g)/((1)/(M)+(1)/(m))=T`

Now, we substitute the values:

`(9.8(m)/(s^2)+0.1(9.8(m)/(s^2)))/((1)/(0.25kg)+(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=(mg-T)/(m)`

Substituting the values:

`a=((0.05kg)(9.8(m)/(s^2))-0.45N)/(0.05kg)`

Solving the operations:

`a=0.8(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(m)/(s^2))`

Solving the operations:

`\Sigma F_(net)=0.24N`

Therefore, the net force is 0.24 N.