The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
Answer:
1.9841256 kg
Explanation:
Given;
Length of the swimming pool = 25.0 ft = 7.62 m ( 1 ft = 0.3048 m )
Width of the swimming pool = 18.5 ft = 5.64 m
Depth of the pool = 9.0 ft =
Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m
now,
Volume of the water in the pool = Length × Width × Depth
or
Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³
also,
1 m³ = 1000 L
thus,
110.2292 m³ = 110229.2 L
also it is given that 18 mg of Cl is added to 1 liter of water
therefore,
In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg
or
= 1.9841256 kg
a. The restoring force in the spring has magnitude
F[spring] = k (0.79 m)
which counters the weight of the mass,
F[weight] = (0.46 kg) g = 4.508 N
so that by Newton's second law,
F[spring] - F[weight] = 0 ⇒ k = (4.508 N) / (0.79 m) ≈ 5.7 N/m
b. Using the same equation as before, we now have
F[weight] = (0.75 kg) g = 7.35 N
so that
(5.7 N/m) x - 7.35 N = 0 ⇒ x = (7.35 N) / (5.7 N/m) ≈ 1.3 m
Answer:
Power, P = 924.15 watts
Explanation:
Given that,
Length of the ramp, l = 12 m
Mass of the person, m = 55.8 kg
Angle between the inclined plane and the horizontal, 
Time, t = 3 s
Let h is the height of the hill from the horizontal,
h = 5.07 m
Let P is the power output necessary for a person to run up long hill side as :
P = 924.15 watts
So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.
we get,