Math / Algebra

Question$\text{Given:} \\ \text{Final velocity } v_f = 0.635 \, \text{m/s} \\ \text{Initial velocity } v_i = 0 \, \text{m/s} \\ \text{Mass } m = 132 \, \text{g} = 0.132 \, \text{kg} \\ \text{Change in distance } \Delta d = 98 \, \text{cm} = 0.98 \, \text{m} \\
\text{Solve for the force of friction } F_f \text{ in two different ways:} \\
\text{1. Using motion:} \\ \text{Use the kinematic equation:} \\ v_f^2 = v_i^2 + 2a\Delta d \\ \text{Solve for acceleration } a \text{ and use } F = ma \text{ to find } F_f. \\
\text{2. Using work/energy/power:} \\ \text{Use the work-energy principle:} \\ \text{Work done by friction } = \Delta \text{Kinetic Energy} \\ \text{Calculate the change in kinetic energy and equate it to the work done by friction to find } F_f.$

Studdy Solution

STEP 1

1. The motion occurs in a straight line.
2. The friction force is constant throughout the motion.
3. There are no other forces acting on the object besides friction.
4. The surface is horizontal (no incline).
5. Air resistance is negligible.

STEP 2

1. Solve for friction force using motion equations.
2. Solve for friction force using work-energy principle.
3. Compare the results.

STEP 3

Let's start by using the motion equations to solve for the friction force.
We'll use the kinematic equation: $v_f^2 = v_i^2 + 2a\Delta d$
Substituting the known values: $(0.635 \text{ m/s})^2 = (0 \text{ m/s})^2 + 2a(0.98 \text{ m})$

STEP 4

Simplify and solve for acceleration:
$0.403225 \text{ m}^2/\text{s}^2 = 1.96a \text{ m}$ $a = \frac{0.403225 \text{ m}^2/\text{s}^2}{1.96 \text{ m}} = 0.2057 \text{ m}/\text{s}^2$
Now that we have the acceleration, we can use Newton's Second Law to find the friction force:
$F_f = ma$ $F_f = (0.132 \text{ kg})(0.2057 \text{ m}/\text{s}^2) = 0.02715 \text{ N}$

STEP 5

Now let's solve for the friction force using the work-energy principle.
The work-energy principle states that the work done by friction equals the change in kinetic energy:
$W_f = \Delta KE$ $F_f \Delta d = \frac{1}{2}m(v_f^2 - v_i^2)$
Let's calculate the right side of the equation first:
$\Delta KE = \frac{1}{2}(0.132 \text{ kg})[(0.635 \text{ m/s})^2 - (0 \text{ m/s})^2]$ $\Delta KE = 0.066 \text{ kg} \cdot 0.403225 \text{ m}^2/\text{s}^2 = 0.0266 \text{ J}$

STEP 6

Now we can equate this to the work done by friction:
$F_f (0.98 \text{ m}) = 0.0266 \text{ J}$
Solving for F_f:
$F_f = \frac{0.0266 \text{ J}}{0.98 \text{ m}} = 0.02715 \text{ N}$

STEP 7

Comparing the results:
1. Using motion equations: F_f = 0.02715 N
2. Using work-energy principle: F_f = 0.02715 N
Both methods yield the same result, confirming our calculations.
The friction force is 0.02715 N.

Was this helpful?

Studdy Solution

STEP 1

1. The motion occurs in a straight line.
2. The friction force is constant throughout the motion.
3. There are no other forces acting on the object besides friction.
4. The surface is horizontal (no incline).
5. Air resistance is negligible.

STEP 2

1. Solve for friction force using motion equations.
2. Solve for friction force using work-energy principle.
3. Compare the results.

STEP 3

Let's start by using the motion equations to solve for the friction force.
We'll use the kinematic equation: $v_f^2 = v_i^2 + 2a\Delta d$
Substituting the known values: $(0.635 \text{ m/s})^2 = (0 \text{ m/s})^2 + 2a(0.98 \text{ m})$

STEP 4

STEP 5

STEP 6

Now we can equate this to the work done by friction:
$F_f (0.98 \text{ m}) = 0.0266 \text{ J}$
Solving for F_f:
$F_f = \frac{0.0266 \text{ J}}{0.98 \text{ m}} = 0.02715 \text{ N}$

STEP 7

Comparing the results:
1. Using motion equations: F_f = 0.02715 N
2. Using work-energy principle: F_f = 0.02715 N
Both methods yield the same result, confirming our calculations.
The friction force is 0.02715 N.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. The motion occurs in a straight line.2. The friction force is constant throughout the motion.3. There are no other forces acting on the object besides friction.4. The surface is horizontal (no incline).5. Air resistance is negligible.

STEP 2

1. Solve for friction force using motion equations.2. Solve for friction force using work-energy principle.3. Compare the results.

STEP 3

STEP 4

STEP 5

STEP 6

Now we can equate this to the work done by friction:Ff(0.98 m)=0.0266 J F_f (0.98 \text{ m}) = 0.0266 \text{ J} Ff​(0.98 m)=0.0266 JSolving for F_f:Ff=0.0266 J0.98 m=0.02715 N F_f = \frac{0.0266 \text{ J}}{0.98 \text{ m}} = 0.02715 \text{ N} Ff​=0.98 m0.0266 J​=0.02715 N

STEP 7

Comparing the results:1. Using motion equations: F_f = 0.02715 N2. Using work-energy principle: F_f = 0.02715 NBoth methods yield the same result, confirming our calculations.The friction force is 0.02715 N.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The motion occurs in a straight line.
2. The friction force is constant throughout the motion.
3. There are no other forces acting on the object besides friction.
4. The surface is horizontal (no incline).
5. Air resistance is negligible.

1. Solve for friction force using motion equations.
2. Solve for friction force using work-energy principle.
3. Compare the results.

Now we can equate this to the work done by friction:
$F_f (0.98 \text{ m}) = 0.0266 \text{ J}$
Solving for F_f:
$F_f = \frac{0.0266 \text{ J}}{0.98 \text{ m}} = 0.02715 \text{ N}$

Comparing the results:
1. Using motion equations: F_f = 0.02715 N
2. Using work-energy principle: F_f = 0.02715 N
Both methods yield the same result, confirming our calculations.
The friction force is 0.02715 N.