A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo
rce diagram and determine the net force. Then calculate the acceleration of the object.
1 answer:
Answer:
Net force on the block is 32 N.
Acceleration of the object is 6.4 m/s².
Explanation:
Let the acceleration of the object be m/s².
Given:
Mass of the block is,
Force of pull is,
Frictional force on the block is,
The free body diagram of the object is shown below.
From the figure, the net force in the forward direction is given as:
Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,
Therefore, the acceleration of the object in the forward direction is 6.4 m/s².
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The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s
The distance between the two objects is
x = x₁ - x₂
= 4t + 161 - 48t + 4t²
x = 4t² - 44t + 161
Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
= 4[ (t - 5.5)² - 5.5² ] + 161
x = 4(t-5)² + 40
Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.
Answer: The distance of the closest approach is 40 m.
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s
The distance between the two objects is
x = x₁ - x₂
= 4t + 161 - 48t + 4t²
x = 4t² - 44t + 161
Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
= 4[ (t - 5.5)² - 5.5² ] + 161
x = 4(t-5)² + 40
Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.
Answer: The distance of the closest approach is 40 m.
Answer:
Explanation:
first ball rolls on the porch by total distance
Then again it will move on horizontal floor
also in vertical direction it will drop down
so we have
so magnitude of net displacement of the ball is given as
Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:
here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute =
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:
where ;
∴
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts