Answer:
v = 2.45 m/s
Explanation:
first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height of cliff = 15 m
Vi = Initial Vertical Velocity = 0 m/s
t = time taken = ?
g = 9.8 m/s²
Therefore,
15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²
t² = (15 m)/(4.9 m/s²)
t = √3.06 s²
t = 1.75 s
Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,
s = vt
where,
s = horizontal distance covered = 4.3 m
v = original horizontal velocity = ?
Therefore,
4.3 m = v(1.75 s)
v = 4.3 m/1.75 s
<u>v = 2.45 m/s</u>
Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules
