What is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 N at an angle of 35° above the horizontal?.
1 answer:
The acceleration of the wagon along the ground is 3.6 m/s².
To solve the problem above, we need to use the formula of acceleration as related to force and mass.
Acceleration: This can be defined as the rate of change of velocity.
⇒ Formula:
Fcos∅ = ma................. Equation 1
⇒ Where:
F = Force ∅ = angle above the horizontal m = mass of the wagon a = acceleration of the wagon
⇒ make a the subject of equation 1
a = Fcos∅/m..................... Equation 2
From the question,
⇒ Given:
⇒ Substitute these values into equation 2
a = 44(cos35°)/10 a = 44(0.8191)/10 a = 3.6 m/s²
Hence, The acceleration of the wagon along the ground is 3.6 m/s²
Learn more about acceleration here: brainly.com/question/9408577
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W = 0.060 J
v_2 = 0.18 m/s
Explanation:
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for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
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