A 5kg block is connected to a 10kg block by a cord on a horizontal, frictionless surface. the 10kg block is?

Question

pulled horizontally by a force of 30 N. what is the tension in the cord connecting them?

Answer 1

loook at a FBD for each block

5kg

T=5*a
where T is the tension and a is the acceleration

10kg

30-T=10*a

since we're solving for T, I will isolate a

(30-T)/10=a
T/5=a

so
30-T=2*T
30=3*T
T=10N

j

Answer 2

it rather is only a friction and dynamics subject. From this factor, something with a million is for the 5 kg block and something with 2 is for the ten kg block. Sum of the forces on m1 interior the path of action (the path of action is advantageous) = m1*a What are the forces on m1? the stress from the cord and the friction. Friction acts against action so it rather is damaging m1*a = T - friction Sum of the forces on m1 perpendicular to the path of action (up is advantageous) = 0 What are the forces on m1 interior the perpendicular path? the traditional stress and weight. Weight acts down, so it rather is damaging 0 = N - m1*g So, N = m1*g Defiintion of friction friction = u*N friction = u*m1*g Plug that returned into the right m1*a = T - u*m1*g T = m1*(a+u*g) Now, via fact the pulley is frictionless, we are able to forget approximately with regard to the torque effects of it. shifting onto the ten kg mass putting... There at the instant are not horizontal forces so we are able to forget approximately approximately that... Sum of the forces interior the vertical path (up is advantageous) = -m2*a... the reason it rather is damaging is via the fact we are assuming that via fact that m1 is accelerating interior a similar path via fact the cord, then m2 must be falling and we reported that up is advantageous. What are the forces on m2 interior the vertical path? stress from the chord and weight. Weight acts down so it rather is damaging. -m2*a = T - m2*g Sub in T from the different equation -m2*a = m1*(a+u*g) - m2*g resolve for 'a' *** tousled previously here ON MY ALGEBRA *** m2*g - m1*u*g = a*(m1+m2) a = (m2*g - m1*u*g) / (m1+m2) Plug on your numbers a = (10 kg * 9.80 one m/s^2 - .30*5kg*9.80 one m/s^2) / (5 kg + 10 kg) a = 5.fifty six m/s^2 Now only plug that into our T equation T = m1*(a+u*g) T = 5 kg * (5.fifty six m/s^2 + .30 * 9.80 one m/s^2) T = 40 two.5 N

Answer 3

take as a system. total mass=15kg,tot. force=30N,acc=30/15=2metre/sec2
draw fbd of 10kg bloc,we can see dat its acc is 2. the tot. force on it should be 20N.but 30N is in forwrd dirn so 10N should be in bckwrd dirn which cn only be supplied by tension. so tension=10N