a body of mass 10kg on a smooth inclined plane is connected to a smooth pulley to a mass of 15kg as shown above, the acceleration of the system is
\(\frac{1}{4}\)g
\(\frac{8}{25}\)g
\(\frac{1}{2}\)g
\(\frac{2}{5}\)g
\(\frac{3}{4}\)g
Explanation
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Looking at the options, you do not have to input the value of g=10m/s².
Considering the pulley system, for mass 10kg;
F=T-mgsin30
(f=ma and T represents the tension in the string)
so, ma=T-mgsin30
10a=T-5g ...... eq. 1
for mass 15kg,
F=mg-T (the weight is not acting on the inclined plane)
ma=mg -T
15a=-T+15g ..... eq. 2
Solving the two eq(s) simultaneously, and eliminating tension to get a;
10a=T-5g
15a=-T+15g
25a=10g (divide both sides by 25)
a= ⅖g
I hope this helps!
My school pls why aren't the questions complete. Past questions have 50 questions but you guys give 45- 48 questions😢😞. Or am I the only one that has this problem.
why not multiply with 10 i took g into consideration and got a 4. if i divided 4/10 it gives 2/5
so what's wrong with that?
Please i need another detailed explanation because i do not understand the one above. Please help
Both the plane and the pulley are smooth, so frictional force is zero, the tension in the rope will be 15g, the correct answer is g
efficiency= L/e ÷ v.r
where load is obviously 15kg and effort is 10kg, then an angle 30°
15/10 ÷1/sin 30
=1.5÷2 = 0.75 or 3/4 g
From Chidinma
efficiency= L/e ÷ v.r
where load is obviously 15kg and effort is 10kg, then an angle 30°
15/10 ÷1/sin 30
=1.5÷2 = 0.75 or 3/4 g
From Chidinma
