Two crates , one with mass 4.00 kg and the other with 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope. Awoman wearing golf shoes(so she can get traction on the ice)pulls horizontally on the 6.00-kg crate with a force F that gives the crate an acceleration of 2.50 m/s2. (a)What is the acceleration of the 4.00- kg crate?(b)Draw a free- body diagram for the 4.00-kg crate. Use that diagram and Newtons second law to find tension T in the rope that connects the two crates. (c) Draw a free-body diagram for 6.00- kg crate. What is the direction of the net force on the 6.00-kg crate? Which is larger in magnitude, force T or force F? (d)Use part (c) and Newton's second law to calculate the magnitude of the force F.
please! may I have this answer for my assignment
A block moves up a 30° incline under the action of certain forces, three of which are shown at the right. F1 is horizontal and of magnitude 40 N. F2 is normal to the plane and of magnitude 20 N. F3 is parallel to the plane and magnitude 30 N. (a) Determine the work done by each force as the block (and point of application of each force) moves 80 cm up the incline. (b) What is the work done by the gravity? (c) Find the total work done on the block excluding the work of frictional force
(a) To find the acceleration of the 4.00 kg crate, you can use Newton's second law which states that the net force applied to an object is equal to the mass of the object multiplied by its acceleration.
Since the two crates are connected by a light rope, they will experience the same tension force (T). Therefore, the net force on the 4.00 kg crate will be the tension force (T) minus any external force:
Net force on 4.00 kg crate = T - F
Since the 4.00 kg crate is on a frictionless surface, there are no other external forces acting on it. Therefore, the net force on the 4.00 kg crate is just the tension force (T):
Net force on 4.00 kg crate = T
Now, we can equate this net force to the mass of the 4.00 kg crate multiplied by its acceleration (a4):
T = 4.00 kg * a4
Simplifying, we find the acceleration of the 4.00 kg crate:
a4 = T / 4.00 kg
(b) To draw a free-body diagram for the 4.00 kg crate, we need to consider all the forces acting on it. Since it is connected to the 6.00 kg crate by a rope and is being pulled horizontally, there are two forces acting on it:
1. Tension force (T): This is the force transmitted from the 6.00 kg crate to the 4.00 kg crate through the rope.
2. Normal force (N): This is the force exerted by the surface of the frozen pond on the 4.00 kg crate, perpendicular to the surface.
The free-body diagram for the 4.00 kg crate will show these two forces:
----------- (T)
| |
| 4.00 kg |
| crate |
| |
-----------
(c) To draw a free-body diagram for the 6.00 kg crate, we need to consider the forces acting on it. Since it is being pulled horizontally by the woman, there are three forces acting on it:
1. Tension force (T): This is the force transmitted from the 4.00 kg crate to the 6.00 kg crate through the rope.
2. Normal force (N): This is the force exerted by the surface of the frozen pond on the 6.00 kg crate, perpendicular to the surface.
3. Applied force (F): This is the force applied horizontally by the woman wearing golf shoes.
The free-body diagram for the 6.00 kg crate will show these three forces:
----------- (T)
| |
| 6.00 kg |
| crate |
| |
-----------
The direction of the net force on the 6.00 kg crate is in the same direction as the applied force (F), since the crate is accelerating in that direction. The magnitude of force T will be equal to force F, as the tension force is transmitted through the rope and will be the same for both crates.
(d) Using Newton's second law, we can calculate the magnitude of the force F. The net force on the 6.00 kg crate is equal to its mass multiplied by its acceleration:
Net force on 6.00 kg crate = m * a
F - T = 6.00 kg * 2.50 m/s^2
Since we know that the magnitude of force T is equal to force F:
F - F = 6.00 kg * 2.50 m/s^2
Simplifying the equation:
0 = 6.00 kg * 2.50 m/s^2
Hence, the magnitude of the force F is 0 N.
If the heavy crate and the light crate are attached,
a) accelerations are equal.
Tension in rope=4kg*acceleration