sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N
from a point 6.80 m above the ground. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground
1 answer:
Answer:
a)-2m/s^2
b)27.2m/s
Explanation:
Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)
W=455N=weight
W=mg
W=455N=weight
The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration
for point b we use the equations of motion with constant acceleration to find the velocity
Where
Vf = final speed
Vo = Initial speed =0
A = acceleration =2m/s
X = displacement =6.8m
Solving
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Answer:
The skater covers a distance of <u>15 m</u> before stopping.
Explanation:
Let the distance traveled before stopping be 'd' m.
Given:
Mass of the skater (m) = 90 kg
Initial velocity of the skater (u) = 12.0 m/s
Final velocity of the skater (v) = 0 m/s (Stops finally)
Coefficient of kinetic friction (μ) = 0.490
Acceleration due to gravity (g) = 9.8 m/s²
Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.
Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.
Frictional force is given as:
Where, 'N' is the normal force acting on the skater. As there is no vertical motion,
∴
Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:
Now, change in kinetic energy is given as:
Therefore, from work-energy theorem,
Hence, the skater covers a distance of 15 m before stopping.