a 12 kg sled is pulled horizontally with a force of 82 N. the coefficent of the kentic friction of the sled is 0.42. draw a FBD of the sled.
how do u find the acceleration of the sled? and what applied force would be required to cause the sled to move at a constant velocity, explain your logic and show all your calculations.
Fs = mg = 12kg * 9.8N/kg = 117.6N = Weight of sled.
Fp = 117.6sin(o) = o. = Force parallel
to plane due to sled.
Fv = 117.6cos(o) = 117.6N. = Force
perpendicular to plane due to sled.
Ff = u*Fv = 0.42 * 117.6 = 49.4N =
Force due to Kinetic friction.
Fn = Fap - Ff = ma,
82 - 49.4 = 12a,
12a = 32.61,
a = 2.72m/s^2.
To draw a free body diagram (FBD) of the sled, we need to consider all the forces acting on it. In this case, the forces include:
1. Gravitational force (mg): It is acting vertically downward and has a magnitude of 12 kg (mass of the sled) multiplied by the acceleration due to gravity (9.8 m/s²).
2. Applied force (F): It is acting horizontally and has a magnitude of 82 N towards the right.
3. Resistance force due to kinetic friction (fk): It is acting in the opposite direction of the applied force and has a magnitude equal to the coefficient of kinetic friction (0.42) multiplied by the weight of the sled.
The FBD of the sled would look like this:
--------------------
| |
| Sled |
| |
-----------------------
↑ mg
↓
← F
↓
← fk
In this FBD, the upward arrow represents the gravitational force (mg), the rightward arrow represents the applied force (F), and the leftward arrow represents the resistance force due to kinetic friction (fk).
To draw a Free Body Diagram (FBD) of the sled, we need to identify all the forces acting on it. Let's list them out:
1. Gravitational force (mg): Since the sled has a mass of 12 kg, we multiply it by the acceleration due to gravity (9.8 m/s^2) to get the gravitational force acting downward. This force is equal to 117.6 N (12 kg * 9.8 m/s^2).
2. Applied force (F): The sled is being pulled horizontally with a force of 82 N. This force is applied in the direction of motion.
3. Kinetic friction force (f): The coefficient of kinetic friction is given as 0.42. To find the friction force, we multiply the coefficient by the normal force. The normal force is equal to the gravitational force acting downward, which is 117.6 N. So, the friction force is 0.42 * 117.6 ≈ 49.39 N. Since the sled is being pulled to the right, the kinetic friction force acts in the opposite direction (to the left).
Now that we have identified the forces, let's draw the FBD of the sled:
-------------
| |
↑
Gravitational force (mg)
↓
-------------
| |
←Applied force (F)→
| |
-------------
← Kinetic friction force (f) →
In this FBD, the sled is represented as a rectangle, and the forces are depicted as arrows. The direction of each arrow indicates the direction in which the force acts. The length of the arrows is proportional to the magnitude of the force.